# Why didn't Reduce give me the right result?

My code:

Clear[r1, r2];
par[r1_, r2_] := r1 r2 / (r1 + r2);
Reduce[{5 par[r2, 32000]/(r1 + par[r2, 32000]) == 1.8 && par[r1, r2] == 5000}, {r1, r2}]


with this result:

I checked it a dozen times, but I can't see why MMA doesn't give me a value for r1. So, like I always do when I'm desperate :-), I quit and restarted MMA. Copied and pasted my input. And ecco!:

Any idea what could have caused the first result?

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the same input giving different answers implies that there is some (invisible) state that's different, so I think all we need here is a seer :) –  acl Aug 6 '12 at 13:52
@acl - maybe a feature request for version 9? :-). It sometimes happens that there are dependencies I've overlooked, but here there's just r1 and r2, and I cleared those! I can't get it. –  stevenvh Aug 6 '12 at 13:55

One possible cause for the error is a previous (and wrong) definition of par[ ]. For example:

par[r1, r2] := 5000;
Clear[r1, r2];
par[r1_, r2_] := r1 r2/(r1 + r2);
Reduce[{5 par[r2, 32000]/(r1 + par[r2, 32000]) == 1.8 &&  par[r1, r2] == 5000}, {r1, r2}]


Edit

 par[r1, r2] := {};


Has the same effect :)

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Looks good, er, I mean bad, you know what I mean. I checked the history of that notebook, and it has a previous definition for par[], but it's basically the same (only written as 1/(1/r1 + 1/r2) ) –  stevenvh Aug 6 '12 at 15:17
@stevenvh Do you have notebooks with a built-in history? You're testing a development version I must assume, since mine hasn't got that capability. I can rewrite an existing line of code leaving no trace of the original one, while the results of its execution are still alive. –  Sjoerd C. de Vries Aug 6 '12 at 15:23
@Sjoerd - I mean, I had saved the notebook before restarting MMA, and since I never delete previous calculations all my IN[]s and Out[]s were still there, only without the references. That's how the first function doesn't show In[] and Out[], while the one after restarting does. Version 8, but that's what MMA has always done, isn't it? –  stevenvh Aug 6 '12 at 15:28
@stevenvh I don't think many mma share your notebook usage habits. Starting a new cell for each and every new calculation even if you only want to re-execute a calculation because you changed some variables? Even if some of the output are high memory density plots? Nah. –  Sjoerd C. de Vries Aug 6 '12 at 19:33
@Verbeia And that solves partially the lacking of multiple undos by enforcing habits and discipline. Now, if I only could remember where I parked my car today ... –  belisarius Aug 8 '12 at 6:20

If you replace the number 1.8 with it's rational value 9/5 you get an answer without warnings:

Clear[r1, r2];
par[r1_, r2_] := r1 r2/(r1 + r2);
Reduce[{5 par[r2, 32000]/(r1 + par[r2, 32000]) == 9/5 &&
par[r1, r2] == 5000}, {r1, r2}]

% // N


With the output:

r1 == 4000000/333 && r2 == 4000000/467
r1 == 12012. && r2 == 8565.31

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This still doesn't answer the question of what caused the first output shown... –  Ｊ. Ｍ. Aug 6 '12 at 13:28
@J.M. I coudn't reproduce the first output. –  Peter Breitfeld Aug 6 '12 at 13:39
@Peter - thanks. Maybe I should have said I'm an engineer, and for engineers the result is 12000 and 8600; we don't care that much about 10 significant digits :-). (For quick calculations in my head pi is often 3. Go figure.) –  stevenvh Aug 6 '12 at 13:52