Half life of a mixture

[bleeder]

I have a real world math problem that's way out of my league. I'm hoping somebody can push me in the right direction.

People with hemophilia have to regularly inject a protein to help their blood clot. As soon as it's injected it begins to decay. The level in the blood at time T is given by: Level=Q*(1/2)^(T/h) . There's a test to measure the level, and it's easy to solve for h:
h=T/log2(Q/Level)

The complication is that many people do multiple injections within a few half life cycles, so the Level is actually a sum:
Level= Q1*(1/2)^(T1/h)+Q2*(1/2)^(T2/h)+Q3*(1/2)^(T3/h)... Now what?

Any suggestions greatly appreciated.

Dan

 

[Deleted member 4993]

I have a real world math problem that's way out of my league. I'm hoping somebody can push me in the right direction.

People with hemophilia have to regularly inject a protein to help their blood clot. As soon as it's injected it begins to decay. The level in the blood at time T is given by: Level=Q*(1/2)^(T/h) . There's a test to measure the level, and it's easy to solve for h:
h=T/log2(Q/Level)

The complication is that many people do multiple injections within a few half life cycles, so the Level is actually a sum:
Level= Q1*(1/2)^(T1/h)+Q2*(1/2)^(T2/h)+Q3*(1/2)^(T3/h)... Now what? <<< What do you need to find??

Any suggestions greatly appreciated.

Dan

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you

 

[bleeder]

In the equation: Level= Q1*(1/2)^(T1/h)+Q2*(1/2)^(T2/h)+Q3*(1/2)^(T3/h), I need to solve for h.

Sorry, I can't show my work because I'm stuck at the beginning. It seems like a straightforward problem, but none of the references I've found have anything like it. I've been out of school for 30 years, so my log rules are pretty rusty.

 

[mmm4444bot]

… Level= Q1*(1/2)^(T1/h) + Q2*(1/2)^(T2/h) + Q3*(1/2)^(T3/h), I need to solve for h …

… It seems like a straightforward problem …

It might not be as straightforward as you expect. I put the equation above into Maple (a computer algebra system), and it solved for h.

Double-click the image below to expand, if necessary.



In the denominator, the symbol _Z is a variable. (You can think of it as x, if you like.)

The notation RootOf() represents all of the values of _Z that cause the expression inside RootOf() to equal zero. You then divide each of those values of _Z into -ln(2)*T3, and that's how many values of h satisfy your original equation.

It could turn out that most of these values of h are Complex numbers (probably ignored, in your application), and only a few are Real numbers. Without knowing the values of the parameters, it's hard to say how many Real values of h there are.

If you post actual values for Level, Q1, Q2, Q3, T1, T2, and T3, then I'll show you the values of h, for those parameters.

 

[bleeder]

Thanks. It's a little comforting to know that the answer isn't something simple like: "just invert and multiply." Does the Maple answer mean that there is not an analytic solution to this problem?

Typical values for Q are between 50 and 100. Or, .5 and 1, they're % of normal figures. T's range from 8 to 96 hours.

It may be beyond the scope of this forum, but what is it that makes this problem so difficult? With a single term the solution is simple- log both sides, eliminate the exponent, solve for h. Add a second term just like the first and you have to fire up the HAL 9000. It's like 1+1 = orange.

Thanks again for your help.

 

[Denis]

> The level in the blood at time T is given by: Level=Q*(1/2)^(T/h) .
> There's a test to measure the level, and it's easy to solve for h:
> h=T/log2(Q/Level)

I make that:
h = T log(1/2) / log(Level / Q)

> Level= Q1*(1/2)^(T1/h) + Q2*(1/2)^(T2/h) + Q3*(1/2)^(T3/h)... Now what?

That's a summation, not necessarily 3 terms only, right?

Since h has to be the same in all terms, why worry: just get it from any of the terms;
like from term#2: h = T2 log(1/2) / log(Level / Q2)
...I'm missing something?

 

[bleeder]

Yes, h is the same in all the terms, but I can only measure the level in the mixture. I'd probably take it out to 5 half life periods, and include all the terms in that time.

 

[Denis]

Well (unless I'm missing something or am way out in left field), I don't think you have
any choice: each needs to be calculated individually.

Could easily be done with a short looper program asking:
how many? input n
1st values? input T1, Q1
...
nth values? input Tn, Qn

 

[mmm4444bot]

… Does the Maple answer mean that there is not an analytic solution to this problem? …

… what is it that makes this problem so difficult? …

My knowledge is not sufficient to give definitive answers to these questions.

As far as software goes, we're always at the mercy of the developers. In other words, it's up to us to properly interpret results from machines. I would guess that the Maple expression for h means there is no straightforward formula for h.

I would guess that the difficulty lies in the fact that h is (1) in the exponent position, (2) spread across sums of terms, and (3) there are no Properties of Logarithms for dealing with (1) and (2) "nicely".

Denis suggested writing code to crunch out a numerical approximation; that's probably the best way to go. If you're handy with Excel (or "sum" such programmable product), then Denis or others (maybe me, even) could probably help you with the code.

You did not supply a value for Level, so I just picked three arbitrary values from the set you gave for Qn and averaged them. Double-click the image, if necessary, to see the resulting values for h. (I drew a green box around the single Real solution, in order to save you eye strain.)

[attachment=0:qbcmsl4t]YukYuk.JPG[/attachment:qbcmsl4t]

 

[Denis]

I do iterations using UBasic; got same results as Mark got from this simple program:

100 Q1=.66:Q2=.89:Q3=.71:T1=15:T2=44:T3=53:L=.7533
110 h=1:i=1
120 k= Q1*(1/2)^(T1/h) + Q2*(1/2)^(T2/h) + Q3*(1/2)^(T3/h)
130 IF k<L THEN h=h+i : GOTO 120
140 h=h-i : i=i/10
150 if accuracy<24 THEN GOTO 120 ELSE PRINT h

21.569 830 939 946 886 811 778 194

BTW, UBasic can go to "1252 accuracy"!

 

[bleeder]

Thanks a bunch guys. I had no idea this problem would be so complex (or imaginary). I really appreciate the help.