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In some textbooks, I have seen the authors treat differentials such as $dx$ as a difference, so $dx$ = $x_2$ - $x_1$

My question is, how legal is this? I realize the differential is a quantity smaller than any conceivable real number but shouldn't this be $dx$ $\approx$ $x_2$ - $x_1$?

Steven
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  • I voted to migrate this to Math.SE. – AccidentalFourierTransform Jun 05 '18 at 18:35
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    The limit of a difference is actually the more rigorous way to think about it, because that's how the differential is defined in the definition of the derivative. An isolated differential doesn't mean anything without another differential on the opposite side of the equation; in that situation, the equation represents a linear approximation for how a difference in one quantity induces a difference in another. – probably_someone Jun 05 '18 at 18:41
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    I posted this for physicists because I wanted their take on the use of differentials. Mathematicians tend to be put off by the non-rigorous use of mathematics that physicists work with. – Steven Jun 05 '18 at 19:32
  • @Steven - That's true. Physicists tend to be more careless than mathematicians with mathematical notations and concepts because they're only interested in using the mathematical tools to get the right answer, even if the procedure that they use may be considered to be sloppy by mathematicians. So, yes, the procedure that the authors of your textbook used may not be strictly "legal" from a mathematician's standpoint. Can't say anything more without knowing exactly how the authors of your textbook used the differentials. –  Jun 05 '18 at 19:57
  • @SamuelWeir

    Here is an example taken from S.A. Elder's book "Fluid Physics for Oceanographers and Physicists":

    "A dry column of the atmosphere has a constant lapse rate from the surface to an altitude of 2000 meters. If the sea level temperature is 0 degrees C, and the temperature at 2000 meters is 30 degrees C, is this air column stable?"

    He then proceeds to calculate the lapse rate as

    $\lambda$ = $\frac{dT}{dh}$ = -(-30-0)/(2000-0)

     – Steven Jun 05 '18 at 22:54
  • @Steven - Well, he did say that the column has a "constant" dT/dh lapse rate so in that case $\Delta T/\Delta h$ = $dT/dh$ exactly and there is no issue with infinitesimal approximations here. –  Jun 05 '18 at 23:41
  • @SamuelWeir

    So would I be correct in assuming that as long as a rate is constant, then $\frac{dT}{dh}$ = $\frac{\Delta T}{\Delta h}$ but if there is a change, regardless of how small, then $\frac{dT}{dH}$ $\approx$ $\frac{\Delta T}{\Delta h}$?

     – Steven Jun 06 '18 at 00:31
  • $\Delta T/\Delta h$= (30-0)/(2000-0) is the average lapse rate between h=0 and h=2000 meters. Unless more information is given, that's all you can say based on the fact that T=0 at h=0 and T=30 C at h=2000 meters. You have no way of knowing how the lapse rate might vary between those two heights. But if you are given the additional information that the lapse rate was constant all the way between these two heights, then you know that the dT/dh throughout this interval has to be equal to the average value of $\Delta T/\Delta h$. –  Jun 06 '18 at 03:52
  • Thank you, Dr. Weir. You've been very helpful and I appreciate you taking the time to help me understand this better. – Steven Jun 06 '18 at 04:31

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