## Leibnitz notation in Calculus

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### Re: Leibnitz notation in Calculus

Now where all this is going is to justify the notational convention (this is not really a proof, just a justification of convention):

this is obviously the sticking point:

stcordova

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### Re: Leibnitz notation in Calculus

But, since this is really an argument about convention, I'm only going to try to justify the convention, but not make a formal proof by going back to that trivial "differential" equation:

multiplying both sides by dx^2:

and then expanding

stcordova

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### Re: Leibnitz notation in Calculus

Now one might complain about the meaning of "d (dy)", but if I may be irreverent to interpret the meaning: Thus double integrating both sides of:

I get

Now, I'm only looking for one solution, not a general solution, so when I do the integrations, I'm going to let the constant of integration C = 0 to make things simpler.
stcordova

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### Re: Leibnitz notation in Calculus

So in light of the idea "(dy)" = "Smiley face":

which I'll carry out without the constants of integration C = 0 just to find a simple solution:

which reduces further to:

double checking that this is one of an infinite number of possible solutions of the "differential" equation:

stcordova

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### Re: Leibnitz notation in Calculus

So going back to the sticking point, is this true, or is the notation at least justified?

In light of the fact we can excute the double integration in any order, it seems to me, we can execulte the differentiation in any order. To make the point a little clearer (I hope anyway):

thus it seems reasonable (though I haven't worked it out more formally) as illustrated in the double integral example where the order integration is interchangeable:

[img]https://latex.codecogs.com/png.latex?\frac%20{d\frac{dy}{dx}}{dz}=\frac%20{d\frac{dy}{dz}}{dx}=\frac{d^2y}{dxdz}[/img]

again this would be (if true) an idealization of:

[img]https://latex.codecogs.com/png.latex?\frac%20{\Delta%20\left%20(%20\frac{\Delta%20y}{\Delta%20x}%20\right%20)}{\Delta%20z}=\frac%20{\Delta%20\left%20(%20\frac{\Delta%20y}{\Delta%20z}%20\right%20)}{\Delta%20x}=\frac{\Delta%20\left%20(%20\Delta%20y%20\right%20)}{\Delta%20x\Delta%20z}[/img]

recall I showed one can double integrate this:

which is an approximation of:

stcordova

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### Re: Leibnitz notation in Calculus

There is probably a formal way to prove this relation as all the delta's go to zero. The problem is that in elementary calculus books, this proof is omitted! For all the proof-based calculus book out there, this was one case where the proof would have been helpful, if only in the appendix or somewhere, rather than just letting the students remain in a state of confusion:

stcordova

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### Re: Leibnitz notation in Calculus

So trying to deal with:

The right hand side of the equation relates to this. One can see the h^2 correspondes to dx^2 (I guess):

The left hand side (I think) relates to:

I think coverting the "h" to "delta x", plus some clean up, will get the desired result to justify the Liebnitz notation.

It would have been helpful if they showed the connection in calculus text rather than just throwing Liebnitz notation in with not a lot of justification.
stcordova

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### Re: Leibnitz notation in Calculus

I realized I went about things the hard way. Cleaning up a bit, and taking out the "z" and replacing with "x".

Note:

this relation has to just be recast to delta-X's and Y's, I think.
stcordova

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### Re: Leibnitz notation in Calculus

Let me recast:

If I define:

and

then

which justifies the notation

stcordova

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### Re: Leibnitz notation in Calculus

Quoting Johnny B:
http://theskepticalforum.org/index.php? ... 380#msg380

Yes, it is a terrible tragedy that this never gets explained in textbooks! What is even more of a tragedy is that, I believe, the notation itself is incorrect. This is detectable if one actually takes seriously as a differential.

To perform this operation, you would actually need to use the quotient rule. Doing so leads to a more expanded form of the second derivative:

Now, this reduces to in the case of x being an independent variable. But, when x is not an independent variable, then the full expansion is needed.

If you have the full expansion, you can easily convert the second derivative of y with respect to x into the second derivative of x with respect to y using algebraic manipulations only, which is not possible with the traditional notation.

and

http://theskepticalforum.org/index.php? ... 400#msg400

Just to jump back in the topic - my paper on an expanded second derivative notation got accepted into the arXiv! http://arxiv.org/abs/1801.09553
stcordova

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