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I posted a link to this topic, so the OP could find my response.Jonah must take an antibiotic every 12 hours. Each pill is 25 milligrams, and after every 12 hours, 50% of the drug remains in his body. What is the amount of antibiotic in his body over the first two days? What amount will there be in his body in the long run?

Here's what I did:

For my recursive formula, I tried inputting u(n)=u(n-1)*(1-0.50). However, I could not seem to find the long run value.,,, It keeps changing. Am I right on this?

I appreciate your help! Thank you so much!

Source: <Advanced Algebra: An Investigative Approach; Murdock Jerald, 2004>

Here is a link to the original question:

Recursive formula help on real world situations? - Yahoo! Answers

Let $A(t)$ represent the amount, in mg, of antibiotic in Jonah's bloodstream at time $t$, measured in hours. With a half-life of 12 hours, we may state:

$\displaystyle A(t)=A_02^{-\frac{t}{12}}$

So, for the first 12 hours, but not including the second dose, we find:

$\displaystyle A_0=25$ hence:

$\displaystyle A(t)=25\cdot2^{-\frac{t}{12}}$

$\displaystyle A(t)=25\cdot2^{-\frac{12}{12}}=\frac{25}{2}$

Now, for the second 12 hours, we have:

$\displaystyle A_0=\frac{25}{2}+25=\frac{75}{2}$ hence:

$\displaystyle A(t)=\frac{75}{2}\cdot2^{-\frac{t}{12}}$

$\displaystyle A(12)=\frac{75}{2}\cdot2^{-\frac{12}{12}}=\frac{75}{4}$

For the third 12 hours, we have:

$\displaystyle A_0=\frac{75}{4}+25=\frac{175}{4}$ hence:

$\displaystyle A(t)=\frac{175}{4}\cdot2^{-\frac{t}{12}}$

$\displaystyle A(12)=\frac{175}{4}\cdot2^{-\frac{12}{12}}=\frac{175}{8}$

For the fourth 12 hours, we have:

$\displaystyle A_0=\frac{175}{8}+25=\frac{375}{8}$ hence:

$\displaystyle A(t)=\frac{375}{8}\cdot2^{-\frac{t}{12}}$

$\displaystyle A(12)=\frac{375}{8}\cdot2^{-\frac{12}{12}}=\frac{375}{16}$

So, now we may try to generalize for the $n$th 12 hour period.

Let $I_n$ represent the initial amount for each 12 hour period. We see we must have the recursion:

$I_{n+1}=\frac{1}{2}I_{n}+25$

This is an inhomogeneous recurrence, so let's employ symbolic differencing to obtain a homogeneous recurrence:

$I_{n+2}=\frac{1}{2}I_{n+1}+25$

Subtracting the former from the latter, we obtain:

$2I_{n+2}=3I_{n+1}-I_{n}$

The characteristic equation is:

$2r^2-3r+1=0$

$(2r-1)(r-1)=0$

Hence, the closed-form will be:

$I_n=k_1\left(\frac{1}{2} \right)^n+k_2$

We may use our data above to determine the parameters $k_i$:

$I_1=k_1\left(\frac{1}{2} \right)^1+k_2=25$

$I_2=k_1\left(\frac{1}{2} \right)^2+k_2=\frac{75}{2}$

or

$\frac{1}{2}\cdot k_1+k_2=25$

$\frac{1}{4}\cdot k_1+k_2=\frac{75}{2}$

Solving this system, we find:

$k_1=-50,k_2=50$ and so we have:

$I_n=-50\left(\frac{1}{2} \right)^n+50=50\left(1+\left(\frac{1}{2} \right)^n \right)$

And so, for the $n$th 12 hour period, we have:

$\displaystyle A_n(t)=I_n2^{-\frac{t}{12}}$

Now, for the long run, which we may take as implying as n grows without bound, we find:

$\displaystyle \lim_{n\to\infty}I_n=50$

and so we find that for the long run the initial amount for a 12 hour period is is 50 mg and the final amount for that period is 25 mg.

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