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Author Topic: Mathematics Help Thread  (Read 216321 times)

Another

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Re: Mathematics Help Thread
« Reply #1365 on: December 10, 2013, 05:33:39 am »

Please be careful with that conclusion. It can approximately hold for a vast array of cases but the reasoning is flawed. When 2 compounds are dissolved in each other the density in general does not have to obey V(x) = (x*Vs+(1-x)*Vw) because of molecular configurations. Most notable example are mixes of ethanol and water which can be much more dense than either of the pure ingredients.
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1366 on: December 10, 2013, 06:03:03 am »

I know that too, but the density increase is generally negligible, and you can't really generalize the molecular configurations into an equation.
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scrdest

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Re: Mathematics Help Thread
« Reply #1367 on: December 10, 2013, 10:19:00 am »

Please be careful with that conclusion. It can approximately hold for a vast array of cases but the reasoning is flawed. When 2 compounds are dissolved in each other the density in general does not have to obey V(x) = (x*Vs+(1-x)*Vw) because of molecular configurations. Most notable example are mixes of ethanol and water which can be much more dense than either of the pure ingredients.

It's an approximation, a rule of thumb kinda thing. If you took any single density from the density table, it would be slightly lower.

We know that an acid mass concentration of x means that every 1 gram of solution has x grams of acid and (1-x) grams of water. Therefore D(x) = V(x)/m = (x*Vs+(1-x)*Vw)/m = x*Ds + (1-x)*Dw. For Dw = 1 and Ds = 1.5, we get D(x) = 1g/cm³*(1+x/2), which is exactly what you wanted to show.

First off, thanks. I'm sending you an internet beer. Except unless you are a robot you aren't able to consume electricity, so it's just going to spill on your memory and make a mess. Sigh.

However, I wonder why did you write that D(x) = V(x)/m. Shouldn't it be m/V(x) instead? Density equals mass per volume, not the inverse. Furthermore, Dw = 1, but Ds cannot be derived from the formula.
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1368 on: December 10, 2013, 05:42:05 pm »

Er, oops. Yeah, it's the other way around, and then you get
D(x) = m/V(x) = m/(x*Vs+(1-x)*Vw), therefore x*Vs+(1-x)*Vw=m/D(x), therefore x/Ds+(1-x)/Dw = 1/D(x), which is linear in 1/Ds and 1/Dw, and given D(0.36) = 1.18g/cm³ and D(0.18) = 1.09g/cm³, we get Ds~=1.67 and Dw~1.01. Note that now there is no longer a linear relation between the mass concentration and the resulting density.
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Re: Mathematics Help Thread
« Reply #1369 on: December 11, 2013, 06:36:34 am »

I think that 1/A=x/B+(1-x)/C  is similar to the concept of "Harmonic average".
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scrdest

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Re: Mathematics Help Thread
« Reply #1370 on: December 11, 2013, 10:26:55 am »

Er, oops. Yeah, it's the other way around, and then you get
D(x) = m/V(x) = m/(x*Vs+(1-x)*Vw), therefore x*Vs+(1-x)*Vw=m/D(x), therefore x/Ds+(1-x)/Dw = 1/D(x), which is linear in 1/Ds and 1/Dw, and given D(0.36) = 1.18g/cm³ and D(0.18) = 1.09g/cm³, we get Ds~=1.67 and Dw~1.01. Note that now there is no longer a linear relation between the mass concentration and the resulting density.

But I'm looking for D(x), not Ds.
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1371 on: December 11, 2013, 10:32:36 am »

Er, oops. Yeah, it's the other way around, and then you get
D(x) = m/V(x) = m/(x*Vs+(1-x)*Vw), therefore x*Vs+(1-x)*Vw=m/D(x), therefore x/Ds+(1-x)/Dw = 1/D(x), which is linear in 1/Ds and 1/Dw, and given D(0.36) = 1.18g/cm³ and D(0.18) = 1.09g/cm³, we get Ds~=1.67 and Dw~1.01. Note that now there is no longer a linear relation between the mass concentration and the resulting density.

But I'm looking for D(x), not Ds.
I simply extrapolated Ds from the data you gave me. If you want D(x) from Ds and Dw, you just put Ds, Dw and x into the formula above: D(x) = 1/(x/Ds+(1-x)/Dw).
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scrdest

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Re: Mathematics Help Thread
« Reply #1372 on: December 11, 2013, 05:17:20 pm »

Er, oops. Yeah, it's the other way around, and then you get
D(x) = m/V(x) = m/(x*Vs+(1-x)*Vw), therefore x*Vs+(1-x)*Vw=m/D(x), therefore x/Ds+(1-x)/Dw = 1/D(x), which is linear in 1/Ds and 1/Dw, and given D(0.36) = 1.18g/cm³ and D(0.18) = 1.09g/cm³, we get Ds~=1.67 and Dw~1.01. Note that now there is no longer a linear relation between the mass concentration and the resulting density.

But I'm looking for D(x), not Ds.
I simply extrapolated Ds from the data you gave me. If you want D(x) from Ds and Dw, you just put Ds, Dw and x into the formula above: D(x) = 1/(x/Ds+(1-x)/Dw).

Uh, I managed to get to that equation myself, actually, and the problem is that Ds is ms/Vs, and since Vs = 0... I tried to use Ds of gaseous HCl (i.e. 36 g/24.4 dm3) and that didn't work (the results were WAY off).
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1373 on: December 11, 2013, 11:22:13 pm »

Well obviously you can't measure the density of something if you have a sample of 0 cm³ of it. How did you expect that to work? Maybe you should just try figuring out the density by taking a different sample size of acid and measuring that, and then use the fact that density is independent of sample size. If you can't figure out the density of pure dissolved HCl you could just, you know, measure it by reversing the formula and applying it to a sample solution of acid (try not using pure water again), then you get D(s) = x/(1/D(x)-(1-x)/Dw).
« Last Edit: December 11, 2013, 11:26:04 pm by MagmaMcFry »
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RedWarrior0

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Re: Mathematics Help Thread
« Reply #1374 on: December 13, 2013, 10:48:05 am »

Red``   my prof showed us some voodoo method of relating two characteristic polynomials
   smeding   yeah i'm out
   Red``   But I think it was if the determinant of E is 0
   Red``   and you can write E = aD^3 + bD^2 + cD + dI or something nonsense like that
   Red``   that the characteristic polynomial of E is going to be ax^3 + bx^2 + cx + d
   Red``   and also that E and D have the same eigenvalues?

I believe this is correct, but I don't quite remember that lecture, it sort of made my skull feel like a sphere that got turned inside-out (which is totally possible, btw). I'm just confirming that I understood it correctly?
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1375 on: December 13, 2013, 02:32:45 pm »

I'm pretty sure that this is false, given that if the determinant of E is 0, the constant coefficient d of its characteristic polynomial would also always be 0. Here's a true statement that sounds kinda similar to your statement: Given a matrix D with the eigenvalues xi and a polynomial p, the eigenvalues of p(D) are exactly p(xi).
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Tomcost

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Re: Mathematics Help Thread
« Reply #1376 on: December 14, 2013, 01:14:43 pm »

I hereby come to this place in seek of an answer to a stupid question, but that I just can't figure out:

limx->-∞ x*ex.

It tends to 0, yet I can't figure out how to apply L'Hopital's rule without getting into an endless loop.

MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1377 on: December 14, 2013, 02:11:07 pm »

Try writing it as x/e^-x.
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Tomcost

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Re: Mathematics Help Thread
« Reply #1378 on: December 14, 2013, 03:37:09 pm »

Oh, I never thought of trying the other way around! Thank you. I guess that it was a pretty stupid mistake, but sometimes one can't just figure out these little things...

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Re: Mathematics Help Thread
« Reply #1379 on: December 14, 2013, 07:07:57 pm »

Anybody know any good guides for how to do trigonometric proofs? (I think that's the name for it).  Or a place that has a large amount of practice problems?
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