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Sequences/Series Problem Based on Medication

  1. Sep 22, 2012 #1
    Hi, I am a sophomore in college and, well I have to do this problem that makes no sense to me... I really don't know where to start. Would you help, and maybe explain how to develop these equations as well?

    It says...

    "For 5 mg some medication, the instructions on the bottle are: Take 8 tablets on Day 1, 7 on Day 2, and decrease by one tablet each day until all tablets are gone. This medication decays exponentially in the body, and 24 hours after taking k mg, there are kx mg in the body.

    a) Write formulas involving x for the amount of medication in the body.
    - 24 hours after taking the first dose (8 tablets), right before taking the second dose (7 tablets).
    - Immediately after taking the second dose. (7 tablets).
    - Immediately after taking the eighth dose (1 tablet)
    - n days after taking the eighth dose.

    b) Find a closed form for the sum T = 8x^7 + 7x^6 + 6x^5 + ... + 2x + 1, which is the number of medication tablets in the body right after taking the eighth dose.

    c) If a patient takes all the medication as prescribed, how many days after taking the eighth dose is there less than 3% of a medication tablet in the patient's body? The half-life of the medication is about 24 hours.

    d) A patient is prescribed n tablets of medication the first day, n-1 the second, and one tablet fewer each day until all tablets are gone. Write a formula that represent T(sub)n, the number of medication tablets in the body right after taking all tablets. Find a closed form for T(sub)n."

    I have tried for hours and am so frustrated.
     
  2. jcsd
  3. Sep 22, 2012 #2

    SammyS

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    Hello mundane. Welcome to PF !

    Integrate T.

    That gives the first eight terms of a geometric series, which converges if 0 < x < 1 .
     
  4. Sep 22, 2012 #3
    For part b I should integrate T? I don't quite understand why
     
  5. Sep 22, 2012 #4

    Ray Vickson

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    Exactly which part do you not understand?

    You start day 1 by taking 8*5 = 40 mg. In terms of x, how much of the day 1 medication is still in the body at the start of day 2? How much of the day 1 medication is still present at the start of day 3? Just keep going like that, and, of course, add new sources due to meds taken at the start of day 2, day 3, etc. I am 100% serious here: if you answer the questions I have asked, you will soon see how to do the whole question.

    BTW: integration has nothing to do with this question, at least as I read it.

    RGV
     
    Last edited: Sep 22, 2012
  6. Sep 22, 2012 #5
    I didn't understand the integration comment.

    For the part a, should I write T1=(8)(5)(x)=40x mg in the body 24 hours after first dose??
    And then T2=40(x)2+(7)(5)=40(x)2+35 for after the second dose?

    Or am I way off? Should I be getting an answer for part a? I just did the x2 bit because it says it decays exponentially but I have no idea if that is right.
     
    Last edited: Sep 22, 2012
  7. Sep 22, 2012 #6
    I REALLY need to understand how to get the closed form for part b. I think I understand part a now, but I don't know where to start with b.
     
  8. Sep 22, 2012 #7

    Ray Vickson

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    Well, just read what the question says: "24 hours after taking k mg, there are kx mg in the body." I guess you need to argue that this exponential decay also applies to meds already in the body---not just to those that are taken---but it is clear enough from the description. Also: you need to be careful about how you measure the dose: just one microsecond before taking the day-2 dose your concentration is 40*x, but one microsecond after taking the day-2 dose it is 40*x + 35. You need to choose which to use. Aside from that, your method looks OK.

    So, what is your problem with part (b)? Don't the terms 8x^7, 7x^6, etc., remind you of something? (Remember: this is Calculus!) I suppose you could legally argue that the expression given in part (b) is already "closed-form", but that would be unwise: the question clearly wants you to come up with a simpler formula.

    RGV
     
  9. Sep 22, 2012 #8
    My problem is that I don't understand the concept of closed form.

    Is it something like Tn=n(xn-1)?

    But that doesn't make sense because n, as I have defined it, is not 8 when you take 8 pills. It is 1.

    I would get this if the number of days weren't going up while the number of pills were going down. That is throwing me for a loop and I don't understand what to do.

    So far I have written:

    T0=(8)(5)=40
    T1=(8)(5)(x)=40x
    T2=40x+35
    T8=40(x)7+35(x)6+...+5.
    Tn=40(x)n-1+35(x)n-2+...+5(x)n-8.

    But in these cases, I am obviously using the amount of days as n. I don't understand how to get closed form for part b) at all. In fact, I don't know how to do b) through d) because I have researched closed form for the past 5 hours and still don't know how I can apply it here, and for part c) I don't know how on earth I would find the point at which there is 3% left if x is just some nebulous number.

    I appreciate your responses, I just really need to see how this thing works because I am literally totally lost. I have been at this for the entire day and I have about 9 hours of homework to finish on top of this by tomorrow night... somebody please just get me started :(
     
    Last edited: Sep 22, 2012
  10. Sep 23, 2012 #9

    SammyS

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    I believe that when I suggested that you integrate T, I also told you why.

    Did you integrate T?

    What did you get?

    Closed form means that you will get some fairly compact expression for T, rather than that longish sum.

    So, if you can find a fairly simple expression for the anti-derivative of T, then then derivative of that expression will give a fairly simple expression for T. Right?

    So, what do you get when you integrate T?
     
  11. Sep 23, 2012 #10
    When I integrate T I get x(x+1)(x+2)(x+4).

    Is that a better answer? Or do I have to write it in a different notation?
     
  12. Sep 23, 2012 #11
    I guess it could also be x^8 + x^7 + x^6 +...+ x.

    Would that be written in summation notation or something?
     
  13. Sep 23, 2012 #12
    I still really don't get c or d. How would I find the point at which there is 3% left if there is no value for x which seems to be the decay constant?!
     
  14. Sep 23, 2012 #13

    Ray Vickson

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    No. The sum x + x^2 + ... + x^8 has a simple final form that you surely must have see many, many times before. If you have forgotten, look up information on the "geometric series".

    RGV
     
  15. Sep 23, 2012 #14

    Ray Vickson

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    Go back and read the question again. It DOES tell you the value of x!

    RGV
     
  16. Sep 23, 2012 #15
    I'm really sorry, I just ldont see where it tells me the value of x... =\
     
  17. Sep 23, 2012 #16
    Is x equal to 1/2 since 24 hours is the half life? Would I set up a half life exp. decay problem to find .03? I just don't get how to set it up with all of this information.
     
  18. Sep 23, 2012 #17

    vela

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    For example, you can show that
    $$1+x+x^2+x^3+\cdots+x^n = \frac{1-x^{n+1}}{1-x}.$$ The righthand side of that equation is what would be called closed form.
     
  19. Sep 23, 2012 #18

    Ray Vickson

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    Yes, x = 1/2.

    Your expression for T was the amount of meds in the body immediately after taking the final pill. You know x so you can figure out the value of T. Now think: what is the amount of meds still in the body n days after taking the last pill? When n = 0 it is T. What is it when n = 1? What about when n = 2? Keep going until the result you get is ≤ 0.03*5 (that is, 3% of one pill's worth). You don't need a trial and error method (although that would work); you ought to be able to express the solution explicitly in a simple formula.

    RGV
     
  20. Sep 23, 2012 #19

    vela

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    You want to be careful about how you label stuff. Let's use the convention that Tn denotes what's in the body (in mg) right after you take the dose for that day. So on day 1, you take 40 mg, so T1 = 40.

    The question then asks how much is in your body right before taking the second dose. You correctly said it would be 40x. Note that "right before taking the second dose" doesn't match our convention for Tn, so it's not equal to T2, which stands for the amount right after you take the second dose, which is, as you wrote, equal to 40x+35.
     
  21. Sep 23, 2012 #20
    How did you get there? How should I show mine since I don't have a 1 but only terms with values of x?
     
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