Sequences/Series Problem Based on Medication

  • Thread starter mundane
  • Start date
In summary, the patient takes 8 tablets on day 1, 7 on day 2, and then one tablet less each day until all tablets are gone. The patient has 8xmg of medication in their body after taking the 8th tablet.
  • #1
mundane
56
0
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.
 
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  • #2
mundane said:
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.
Hello mundane. Welcome to PF !

Integrate T.

That gives the first eight terms of a geometric series, which converges if 0 < x < 1 .
 
  • #3
SammyS said:
Hello mundane. Welcome to PF !

Integrate T.

That gives the first eight terms of a geometric series, which converges if 0 < x < 1 .

For part b I should integrate T? I don't quite understand why
 
  • #4
mundane said:
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.

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
 
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  • #5
Ray Vickson said:
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

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.
 
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  • #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.
 
  • #7
mundane said:
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.

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
 
  • #8
Ray Vickson said:
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

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 :(
 
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  • #9
mundane said:
For part b I should integrate T? I don't quite understand why

mundane said:
I didn't understand the integration comment.

...

SammyS said:
...
Integrate T.

That gives the first eight terms of a geometric series, which converges if 0 < x < 1 .
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?
 
  • #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?
 
  • #11
I guess it could also be x^8 + x^7 + x^6 +...+ x.

Would that be written in summation notation or something?
 
  • #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?!
 
  • #13
mundane said:
I guess it could also be x^8 + x^7 + x^6 +...+ x.

Would that be written in summation notation or something?

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
 
  • #14
mundane said:
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?!

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

RGV
 
  • #15
Ray Vickson said:
Go back and read the question again. It DOES tell you the value of x!

RGV

I'm really sorry, I just ldont see where it tells me the value of x... =\
 
  • #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.
 
  • #17
mundane said:
My problem is that I don't understand the concept of closed form.
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.
 
  • #18
mundane said:
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.

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
 
  • #19
mundane said:
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.
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.
 
  • #20
vela said:
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.

How did you get there? How should I show mine since I don't have a 1 but only terms with values of x?
 
  • #21
Ray Vickson said:
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

I think I understand that I should use the half life formula, but I DONT KNOW WHAT to plug in for my initial amount in the Ce^(kt) equation if that is what I use
 
  • #22
Guys, I am really lost and have never been this frustrated/felt this stupid in my life. Can I just see what I should do so I can do the rest of these problems?
 
  • #23
For b, when I integrate T and give the closed form sum, do I write my answer in terms of (integral)T or say that T is the derivative of that result?


I currently have ∫Tdx=x8+x7+...+x=xn, so T=(dy/dx)(xn)

is that good?
 
  • #24
Please respond I am about to freaking bawl.
 
  • #25
Is there a common ratio in this series?
 
  • #26
Somebody please help.
 
  • #27
mundane said:
For b, when I integrate T and give the closed form sum, do I write my answer in terms of (integral)T or say that T is the derivative of that result?


I currently have ∫Tdx=x8+x7+...+x=xn, so T=(dy/dx)(xn)

is that good?
That's a start. But it's not equal to xn, and I wouldn't call it xn, if that's what you had in mind.

Let's define the sum Sn as [itex]\displaystyle S_n=x+x^2+x^3+\dots+x^n=\sum_{k=1}^{n}x^k\ .[/itex]

Then [itex]\displaystyle \int T\,dx=S_8\,,[/itex] so [itex]\displaystyle T=\frac{d}{dx}(S_8)\ ,[/itex] noting that both T & S8 are functions of x.

Now let's help you find a closed form for S8.

Extending the definition of Sn, let [itex]\displaystyle S_\infty=x+x^2+x^3+\dots=\sum_{k=1}^{\infty}x^k\ .[/itex] S does converge, for 0 < x < 1, which is the case that you have.

Now for a trick:
[itex]\displaystyle S_\infty=x + x(x+x^2+x^3+\dots )[/itex]
[itex]\displaystyle =x+x\cdot S_\infty\ .[/itex]

Solve for S .

Now notice that:

[itex]\displaystyle S_8=S_\infty-x^9+x^{10}+x^{11}+\dots[/itex]
[itex]\displaystyle =S_\infty-x^8\cdot S_\infty\ .[/itex]​

Factor, then plug-in the expression you have for S to find S8, then find T.
 
  • #28
SammyS said:
That's a start. But it's not equal to xn, and I wouldn't call it xn, if that's what you had in mind.

Let's define the sum Sn as [itex]\displaystyle S_n=x+x^2+x^3+\dots+x^n=\sum_{k=1}^{n}x^k\ .[/itex]

Then [itex]\displaystyle \int T\,dx=S_8\,,[/itex] so [itex]\displaystyle T=\frac{d}{dx}(S_8)\ ,[/itex] noting that both T & S8 are functions of x.

Now let's help you find a closed form for S8.

Extending the definition of Sn, let [itex]\displaystyle S_\infty=x+x^2+x^3+\dots=\sum_{k=1}^{\infty}x^k\ .[/itex] S does converge, for 0 < x < 1, which is the case that you have.

Now for a trick:
[itex]\displaystyle S_\infty=x + x(x+x^2+x^3+\dots )[/itex]
[itex]\displaystyle =x+x\cdot S_\infty\ .[/itex]

Solve for S .

Now notice that:

[itex]\displaystyle S_8=S_\infty-x^9+x^{10}+x^{11}+\dots[/itex]
[itex]\displaystyle =S_\infty-x^8\cdot S_\infty\ .[/itex]​

Factor, then plug-in the expression you have for S to find S8, then find T.

I don't understand anything you just did at all. Is this really the process I should be taking?



And for the 3% remaining question, will you just show me how to set it up? I REALLY just need to SEE how it is done. I have another 95 of these to do, can you just show me how to do one of them? I am not going to learn if I just see more language that I don't understand. Please, I am at the end of my road here about to throw my computer out the window and cry.
 
  • #29
For the 3% remaining problem I did this:

T=3.921 pills remaining, so

.03=3.921(1/2t/1 day=3.921t

log(1/2)(t)=log(.03/3.921)=log(.007)
log(1/2)(t)=-2.1163
-.301t=-2.1163
t=7.031 days (round up?)

Is this moderately close? At least tell me that ;'(
 
  • #30
mundane said:
I don't understand anything you just did at all. Is this really the process I should be taking?



And for the 3% remaining question, will you just show me how to set it up? I REALLY just need to SEE how it is done. I have another 95 of these to do, can you just show me how to do one of them? I am not going to learn if I just see more language that I don't understand. Please, I am at the end of my road here about to throw my computer out the window and cry.
Well, it shouldn't be all that hard to follow, but a more direct method came to mind. It just require that you recognize that [itex]\displaystyle (1-x)(1+x+x^2+x^3+\dots
+x^{m-2}+x^{m-1})=1-x^m\ .[/itex]

As it turns there is the constant of integration that was left out of your integration -- or set to zero. You can make it whatever you want, because you take the derivative to get back to T.

[itex]\displaystyle \int T\, dx=C+x+x^2+x^3+\dots+x^7+x^8\ . [/itex]

Choosing C = 1 gives

[itex]\displaystyle \int T\, dx=1+x+x^2+x^3+\dots+x^7+x^8\ . [/itex]
[itex]\displaystyle =\frac{(1-x)(1+x+x^2+x^3+\dots+x^7+x^8)}{1-x}[/itex]

[itex]\displaystyle =\frac{(1-x^9)}{1-x}\ .[/itex]​
Take the derivative of that to get an alternative expression for T. (The resulting expression is not exactly a piece of to evaluate for a specific value of x, but it is a bit easier than using the original expression for T.)
 
  • #31
mundane said:
For the 3% remaining problem I did this:

T=3.921 pills remaining, so

.03=3.921(1/2t/1 day=3.921t

log(1/2)(t)=log(.03/3.921)=log(.007)
log(1/2)(t)=-2.1163
-.301t=-2.1163
t=7.031 days (round up?)

Is this moderately close? At least tell me that ;'(
I get that T = 3.921875 pills ≈ 3.922 pills , so you're probably close enough for that answer.

After, the 8th dose, the patient takes no more pills. The half-life of the pills is approximately 24 hours.
.03=3.921((1/2)t) is correct for t measured in days, but 3.921(1/2)t ≠ 3.921t . (Probably a typo, b/c the result is very good.)

Yes, I concur with your answer..
 
  • #32
mundane said:
For the 3% remaining problem I did this:

T=3.921 pills remaining, so

.03=3.921(1/2t/1 day=3.921t

log(1/2)(t)=log(.03/3.921)=log(.007)
log(1/2)(t)=-2.1163
-.301t=-2.1163
t=7.031 days (round up?)

Is this moderately close? At least tell me that ;'(

How did you get T = 3.921? We have T = 5 + 10*x + 15*x^2 + ... + 40*x^7, where x = 1/2. Just taking the first two terms already gives you 10, and the other terms will give you still more. The rest of you calculation makes no sense either. Why would you say that 3.921*1/2t is the same as 3.921t? (It isn't.) Do you honestly believe that, or are you just writing down random things out of desperation?

RGV
 
  • #33
Ray Vickson said:
How did you get T = 3.921? We have T = 5 + 10*x + 15*x^2 + ... + 40*x^7, where x = 1/2. Just taking the first two terms already gives you 10, and the other terms will give you still more. The rest of you calculation makes no sense either. Why would you say that 3.921*1/2t is the same as 3.921t? (It isn't.) Do you honestly believe that, or are you just writing down random things out of desperation?

RGV
Ray, T is measured in "number of medication tablets", so it's 1/5 of what you have.

The line after the one with 3.921t, indicates that 3.921t was a typo, or was ignored in the next step.
 
Last edited:
  • #34
mundane,

Have you done part (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 represents Tn , the number of medication tablets in the body right after taking all tablets. Find a closed form for Tn .​
 
  • #35
Awesome, thanks, SammyS!

For Tn, in part d) the equation I have is nxn-1+(n-1)xn-2 ... + but I don't know what to end it with. Do I call the number of days y and incorporate that in? I am trying to make it summation notation but can't get anywhere,
 

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