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Mathematics 22 Online
OpenStudy (anonymous):

A patient is prescribed 50mg of medication to be taken every morning for the rest of their life "forever". The physiology of the medication is such that the body metabolizes it so that at the end of a 24 hour period there is only 10% of the medication left in the body. So, on the first morning (day 0) the patient takes 50mg and has 50mgs in their body. the next morning (day 1) there is 5mg left and they take another 50mg. At that moment there is 55mg in their body. The next morning (day 2) there is 5.5 mg left in their body and another 50 mg dose is taken giving 55.5 mg in the body.

OpenStudy (anonymous):

QUESTION: As the days "go to infinity" what is the maximum amount of medication that will ever be in the patients body?

OpenStudy (anonymous):

I've worked out this formula:\[\sum_{d=0}^{\infty}50(1/10)^{d}\]

OpenStudy (anonymous):

And am pretty sure it is describing the phenomenon: http://www.wolframalpha.com/input/?i=+sum+50%281%2F10%29^d%2C+d%3D0+to+0

OpenStudy (anonymous):

hmmim thinkin it has to b at least 55. cuz at the end ur gonna end up with 55 and5/9 but i havent worked out the formula yet

OpenStudy (anonymous):

55.555 is d=4 in the series

OpenStudy (anonymous):

ur series looks correct, i just dont like the answer wolframalpha gave u

OpenStudy (anonymous):

Th sum at least

OpenStudy (anonymous):

Its the sum

OpenStudy (anonymous):

o i kno u ddnt put the summation to infinity. u left it at zero

OpenStudy (anonymous):

So at d=0 we have: 50(1/10)^0 -> 50(1) = 1

OpenStudy (anonymous):

ah yea

OpenStudy (anonymous):

If I put infinity in it says \[\approx 55.556\]

OpenStudy (anonymous):

so at the end u should get 55.5repeating which is 55 +5/9

OpenStudy (anonymous):

I wanna know why though

OpenStudy (anonymous):

Do I have to take a limit in some way?

OpenStudy (anonymous):

well u solved it by doing a summation so u would just have to solve ur summation

OpenStudy (anonymous):

Its weird though

OpenStudy (anonymous):

Wouldnt the amount just keep increasing by a tiny .5ish amount?

OpenStudy (anonymous):

actually each day the amount increases by 5/(10^n). so each day u would add a 5 after the last 1 so the 5 would just grow longer and longer into infinity. 55.5 55.55 55.555 55.5555 55.55555 and so on

OpenStudy (anonymous):

So wouldnt the max amount be 55.56?

OpenStudy (anonymous):

If you are going to approximate anyway on 55.5555556 like wolfram does

OpenStudy (anonymous):

Since I am looking for the max it is safe to just say 55.56?

OpenStudy (anonymous):

where are u writing ur answer. the max amount is 500/9 if u can write that. u cant really write the max amount as a decimal

OpenStudy (anonymous):

Why is that the max amount though?

OpenStudy (anonymous):

It seems arbitrary to me...

OpenStudy (anonymous):

Is there some method you are doin to get that exact amount?

OpenStudy (anonymous):

i solved ur summation. (1/10)^d is an r series

OpenStudy (anonymous):

I guess I dont know how to solve summations?

OpenStudy (anonymous):

if u solve it, it becomes 10/9 and then u multiply that times 50 so u get 500/9

OpenStudy (anonymous):

do u kno the r series?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

I have learned test to determine convergence or divergence and am learning power series.

OpenStudy (anonymous):

summation from o to infinity of r^n=r/(1-r) if 0<r<1

OpenStudy (anonymous):

Not familiar

OpenStudy (anonymous):

geometric series!! u should have learnt that before u learned tests for convergence and divergence

OpenStudy (anonymous):

Summer class lol

OpenStudy (anonymous):

I am looking up "how to solve an infinite summation"

OpenStudy (anonymous):

try looking up geometric series

OpenStudy (anonymous):

Is it very involved or just a few steps?

OpenStudy (anonymous):

its actually quite simple. just plug in the numbers

OpenStudy (anonymous):

\[\sum_{o}^{\infty}r=r/(1-r)\]

OpenStudy (anonymous):

but it only works if: 0<r<1

OpenStudy (anonymous):

Ok thanks for the help!

OpenStudy (anonymous):

np

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