My Math Forum An Aspirin a Day

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 December 14th, 2011, 03:29 PM #1 Newbie   Joined: Dec 2011 Posts: 5 Thanks: 0 An Aspirin a Day After 38 years as an Electrical Engineer, I should be able to solve this but I am having trouble. I take a 325 mg aspirin, then the next day an 81 mg, then the next day a 325 mg and so on for the whole year. Aspirin has a half life of four days in a human. I have to do this for my whole life due to a stroke. My question is, what is the aspirin level in my bloodstream after 30 days and after 365 days? Also, if I were to change to an 81 mg aspirin every day, what would the level be after the 30 and 365 day time frames? Thanks to anyone that can help (By the way, the Doctor had no idea of the answer)
 December 14th, 2011, 06:52 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: An Aspirin a Day $325,\,t\,=\,0$ $325\,\cdot\,$$\frac{1}{2}$$^{\small{t/4}}\,+\,81,\,t\,=\,1$ $325\,\cdot\,$$\frac{1}{2}$$^{\small{t/4}}\,+\,81\,\cdot\,$$\frac{1}{2}$$^{\small{\frac{t \,-\,1}{4}}}\,+\,325,\,t\,=\,2$ $325\,\cdot\,$$\frac{1}{2}$$^{\small{t/4}}\,+\,81\,\cdot\,$$\frac{1}{2}$$^{\small{\frac{t \,-\,1}{4}}}\,+\,325\,\cdot\,$$\frac{1}{2}$$^{\small{ \frac{t\,-\,2}{4}}}\,+\,81,\,t\,=\,3$ $325\,\cdot\,$$\frac{1}{2}$$^{\small{t/4}}\,+\,81\,\cdot\,$$\frac{1}{2}$$^{\small{\frac{t \,-\,1}{4}}}\,+\,325\,\cdot\,$$\frac{1}{2}$$^{\small{ \frac{t\,-\,2}{4}}}\,+\,81\,\cdot\,$$\frac{1}{2}$$^{\small{\ frac{t\,-\,3}{4}}}\,+\,325,\,t\,=\,4$ $325\,\cdot\,$$\frac{1}{2}$$^{\small{t/4}}\,+\,81\,\cdot\,$$\frac{1}{2}$$^{\small{\frac{t \,-\,1}{4}}}\,+\,325\,\cdot\,$$\frac{1}{2}$$^{\small{ \frac{t\,-\,2}{4}}}\,+\,325\,\cdot\,$$\frac{1}{2}$$^{\small{ \frac{t\,-\,3}{4}}}\,+\,325\,\cdot\,$$\frac{1}{2}$$^{\small{ \frac{t\,-\,4}{4}}}\,+\,81,\,t\,=\,5$ $325\,\cdot\,$$\frac{1}{2}$$^{\small{t/4}}\,+\,81\,\cdot\,$$\frac{1}{2}$$^{\small{\frac{t \,-\,1}{4}}}\,+\,325\,\cdot\,$$\frac{1}{2}$$^{\small{ \frac{t\,-\,2}{4}}}\,+\,81\,\cdot\,$$\frac{1}{2}$$^{\small{\ frac{t\,-\,3}{4}}}\,+\,325\,\cdot\,$$\frac{1}{2}$$^{\small{ \frac{t\,-\,4}{4}}}\,+\,81\,\cdot\,$$\frac{1}{2}$$^{\small{\ frac{t\,-\,5}{4}}}\,+\,325,\,t\,=\,6$ These can be worked into two geometric series, both of which converge as t grows without bound. If my calculations are correct, (and this applies to my previous statement) you end up with about 1.3 grams in your system after thirty days and it pretty much stays that way, no matter how long you take it.
 December 14th, 2011, 10:38 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: An Aspirin a Day Let A(t) represent the amount, in mg, of aspirin in your bloodstream at time t, measured in days. With a half-life of 4 days, we may state: $A(t)=A_02^{\small{-\frac{t}{4}}}$ On the first day, we have $A_0=325\text{ mg}$ When you take the 81 mg dose the next day, then adding this to the amount still in your bloodstream from the days before is: $325\cdot2^{\small{-\frac{1}{4}}}+81\approx354.2913349574572$ On the 3rd day we have: $325\cdot2^{\small{-\frac{1}{2}}}+81\cdot2^{\small{-\frac{1}{4}}}+325\approx622.9223135211788$ 4th day: $325\cdot2^{\small{-\frac{3}{4}}}+81\cdot2^{\small{-\frac{1}{2}}}+325\cdot2^{\small{-\frac{1}{4}}}+81\approx604.8131404215097$ It appears we may say: the amount on the nth odd day is: $325\sum_{k=0}^{n-1}$$\frac{1}{2}$$^{\frac{k}{2}}+81\sum_{k=1}^{n-1}$$\frac{1}{2}$$^{\frac{2k-1}{4}}=325$$\frac{2-2^{\frac{2-n}{2}}}{2-2^{\frac{1}{2}}}$$+81$$\frac{1-2^{\frac{1-n}{2}}}{2^{\frac{1}{4}}-2^{-\frac{1}{4}}}$$$ the amount on the nth even day is: $81\sum_{k=0}^{n-1}$$\frac{1}{2}$$^{\frac{k}{2}}+325\sum_{k=1}^{n}\ (\frac{1}{2}\)^{\frac{2k-1}{4}}=81$$\frac{2-2^{\frac{2-n}{2}}}{2-2^{\frac{1}{2}}}$$+325$$\frac{1-2^{-\frac{n}{2}}}{2^{\frac{1}{4}}-2^{-\frac{1}{4}}}$$$ Continuing, and rounding to two places, we find: Day.....A(t) 005.....833.59 006.....781.96 007.....982.55 008.....907.22 009.....1087.88 010.....995.79 011.....1162.36 012.....1058.42 013.....1215.02 014.....1102.71 015.....1252.26 016.....1134.02 017.....1278.60 018.....1156.17 019.....1297.22 020.....1171.83 021.....1310.38 022.....1182.90 023.....1319.69 024.....1190.73 025.....1326.28 026.....1196.26 027.....1330.93 028.....1200.18 029.....1334.22 030.....1202.94 Well before 1 year, the amounts will have settled to: Odd days (right after 325 mg dose): $\lim_{n\to\infty}$$325\(\frac{2-2^{\frac{2-n}{2}}}{2-2^{\frac{1}{2}}}$$+81$$\frac{1-2^{\frac{1-n}{2}}}{2^{\frac{1}{4}}-2^{-\frac{1}{4}}}$$\)=\frac{650}{2-\sqrt{2}}+\frac{81}{2^{\frac{1}{4}}-2^{-\frac{1}{4}}}\approx1342.17$ Even days (right after 81 mg dose): $\lim_{n\to\infty}$$81\(\frac{2-2^{\frac{2-n}{2}}}{2-2^{\frac{1}{2}}}$$+325$$\frac{1-2^{-\frac{n}{2}}}{2^{\frac{1}{4}}-2^{-\frac{1}{4}}}$$\)=\frac{162}{2-\sqrt{2}}+\frac{325}{2^{\frac{1}{4}}-2^{-\frac{1}{4}}}\approx1209.63$ Now, if you decide to take an 81 mg dose every day, then on day n, we would have: $A(n)=81\sum_{k=0}^{n-1}$$\frac{1}{2}$$^{\frac{k}{4}}=81$$\frac{2-2^{\frac{4-n}{4}}}{2-2^{\frac{3}{4}}}$$$ After 30 days, the amount would be about 506.29 mg. The amount would settle to: $81\lim_{n\to\infty}$$\frac{2-2^{\frac{4-n}{4}}}{2-2^{\frac{3}{4}}}$$=\frac{162}{2-2^{\frac{3}{4}}}\approx509.10$
 December 15th, 2011, 06:06 AM #4 Newbie   Joined: Dec 2011 Posts: 5 Thanks: 0 Re: An Aspirin a Day Thank you so much for taking the time to solve my question. Now I have some real data to discuss with my doctor. Again, thank you both for the solutions. Mark
 December 15th, 2011, 01:10 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: An Aspirin a Day I should have looked this up first, but the data I found indicates a half-life of about 3-4 hours for the doses considered here. This changes the problem, and I will work it again this evening using a variable to represent the half-life in hours.
 December 15th, 2011, 06:26 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: An Aspirin a Day Let's say the half-life of aspirin is H hours, and you will alternate between two doses, A taken on the first day and B taken on the second day, and alternating every day after. The doses will be taken exactly 24 hours apart. Then we have: The amount on the nth odd day is: $A\sum_{k=0}^{n-1}$$\frac{1}{2}$$^{\frac{48k}{H}}+B\sum_{k=1}^{n-1}$$\frac{1}{2}$$^{\frac{24$$2k-1$$}{H}}=\frac{A$$2^{\frac{48}{H}}-2^{\frac{48\(1-n$$}{H}}\)+B$$2^{\frac{24}{H}}-2^{\frac{24\(3-2n$$}{H}}\)}{2^{\frac{48}{H}}-1}$ The amount on the nth even day is: $A\sum_{k=1}^{n}$$\frac{1}{2}$$^{\frac{24$$2k-1$$}{H}}+B\sum_{k=0}^{n-1}$$\frac{1}{2}$$^{\frac{48k}{H}}=\frac{A$$2^{\fra c{24}{H}}-2^{\frac{24\(1-2n$$}{H}}\)+B$$2^{\frac{48}{H}}-2^{\frac{48\(1-n$$}{H}}\)}{2^{\frac{48}{H}}-1}$ The odd day amount converges to: $\lim_{n\to\infty}$$\frac{A\(2^{\frac{48}{H}}-2^{\frac{48\(1-n$$}{H}}\)+B$$2^{\frac{24}{H}}-2^{\frac{24\(3-2n$$}{H}}\)}{2^{\frac{48}{H}}-1}\)=\frac{2^{\frac{24}{H}}$$A\cdot2^{\frac{24}{H} }+B$$}{2^{\frac{48}{H}}-1}$ The even day amount converges to: $\lim_{n\to\infty}$$\frac{A\(2^{\frac{24}{H}}-2^{\frac{24\(1-2n$$}{H}}\)+B$$2^{\frac{48}{H}}-2^{\frac{48\(1-n$$}{H}}\)}{2^{\frac{48}{H}}-1}\)=\frac{2^{\frac{24}{H}}$$A+B\cdot2^{\frac{24}{ H}}$$}{2^{\frac{48}{H}}-1}$ If you decide to take the same dose C every day, then we have the amount on day n as: $C\sum_{k=0}^{n-1}$$\frac{1}{2}$$^{\frac{24k}{H}}=C$$\frac{2^{\fra c{24}{H}}-2^{\frac{24\(1-n$$}{H}}}{2^{\frac{24}{H}}-1}\)$ The amount converges to: $C\lim_{n\to\infty}$$\frac{2^{\frac{24}{H}}-2^{\frac{24\(1-n$$}{H}}}{2^{\frac{24}{H}}-1}\)=C$$\frac{2^{\frac{24}{H}}}{2^{\frac{24}{H}}-1}$$$ I'm thinking it might be possible for your doctor to determine a good approximation for H for you through blood tests (from what I read this varies from person to person), and this would allow you to plot the amount in your bloodstream with even more accuracy. I wish the best for you and hope your recovery goes well! -Mark
 December 16th, 2011, 07:10 AM #7 Newbie   Joined: Dec 2011 Posts: 5 Thanks: 0 Re: An Aspirin a Day I was just going to repost when I saw your latest post. Yes, I also did a search and find that the silicylate part has a half life of 2 to 6 hours. I talked to the Neurologist today. He said he misspoke and meant 4 hours, not 4 days. My stroke was 2 years ago. My wife got me to the ER at the first sign and they gave me an ejection of a new drug that disolved the clot. No permanent damage! SOAPBOX: If you ever have something wrong with your face, your arm and your speech, take aspirin and get to the hospital as fast as you can. The doctor was adament and said to keep taking the 325/81 regimine that he prescribed. He did say that 2 of the 81mgs would be Ok, but only if there were some symptoms due to the 325/81. That being said, could you please calculate the 30 day residual, since I am sure the data will truncate within that period. I guess I should have googled the half life prior to posting, but a doctor that gets \$450 per hour should not misspeak.
 December 16th, 2011, 03:29 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: An Aspirin a Day With H = 4 hours, A = 325 mg, and B = 81 mg, we have: 15th odd day (day 29): $\frac{325$$2^{12}-2^{-168}$$+81$$2^6-2^{-162}$$}{2^{12}-1}\approx326.35$ 15th even day (day 30): $\frac{325$$2^6-2^{-174}$$+81$$2^{12}-2^{-168}$$}{2^{12}-1}\approx86.10$ The limiting values are: odd days (after 325 mg dose): $\frac{2^6$$325\cdot2^6+81$$}{2^{12}-1}\approx326.35$ even days (after 81 mg dose): $\frac{2^6$$325+81\cdot2^6$$}{2^{12}-1}\approx86.10$ You were right, the amount did converge nearly to the limiting amount within 30 days. With a half-life of 4 hours, only 1/64 remains after 24 hours. Glad to hear your recovery went well and kudos to your wife for acting so quickly!!
 December 16th, 2011, 05:16 PM #9 Newbie   Joined: Dec 2011 Posts: 5 Thanks: 0 Re: An Aspirin a Day Thank you. I was thinking the solution would be an increasing amount over time, but that does not appear to be the case. Mark (PSL, FL)
 December 16th, 2011, 06:00 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: An Aspirin a Day They are the sum of 2 infinite geometric series which converge to a finite value. To find how quickly they converge to two decimal places, we could use: $\|2^{-6n}-2^{6(1-n)}\|\le0.005$$2^6-1$$$ $2^{-6n}$$2^6-1$$\le0.005$$2^6-1$$$ $2^{-6n}\le0.005$ $n=\frac{1}{6}\log_2(200)\approx1.27$ They will have converged to 2 decimal places by the 2nd iteration. For a variable half-life H, we would have: $n=\frac{H}{24}\log_2(200)$ As we can see, n varies linearly with H. With a half life of 4 days, it takes 31 iterations.

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