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 January 11th, 2017, 11:23 AM #21 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,439 Thanks: 2114 Math Focus: Mainly analysis and algebra I think that's more vague handwaving in an attempt to avoid a circular definition. They clearly aren't ordinary variables as highlighted by the OP. Last edited by v8archie; January 11th, 2017 at 12:00 PM.
January 11th, 2017, 11:51 AM   #22
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Quote:
 Originally Posted by zylo Differentials of a function are defined by equation of tangent line or plane at a point. As such, they can be treated as ordinary variables, including cancellation. When small, they approximate the function in the neighborhood of the point.
That is a precise, mathematical, definition.

Sorry you didn't understand some of my other posts which specifically addressed the OP, where I distinguish between the differential of u(x,y,z), and the differential of u(x,y,z) with two of the variables held fixed, in direct response to OP question why isn't 3$\displaystyle \partial u/\partial u$=3, noting that \partial u is not a differential.

January 11th, 2017, 12:05 PM   #23
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Quote:
 Originally Posted by zylo That is a precise, mathematical, definition.
The fact that you don't understand what a precise mathematical definition is makes most of what you write at best unreliable and at worst unintelligible.

As I said before, I'd encourage the reader to focus on what other people write.

January 11th, 2017, 06:55 PM   #24
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 Originally Posted by photogo01 I ran into the following formula while reviewing the covariant/contravariant base vectors: (âˆ‚x/âˆ‚u)(âˆ‚u/âˆ‚x) + (âˆ‚y/âˆ‚u)(âˆ‚u/âˆ‚y) + (âˆ‚z/âˆ‚u)(âˆ‚u/âˆ‚z) = âˆ‚u/âˆ‚u = 1 Can someone walk me through the first step here that gives "âˆ‚u/âˆ‚u". Why isn't it that you cancel the differentials âˆ‚x, âˆ‚y, and âˆ‚z which leaves you with: âˆ‚u/âˆ‚u+âˆ‚u/âˆ‚u+âˆ‚u/âˆ‚u = 3âˆ‚u/âˆ‚u = 3?
The equation is meaningless and stems from a mindless application of the chain rule. But it's tricky because it looks OK, which forced me to carefully review differentials.

If u =u(x,y),
$\displaystyle du=\frac{\partial u}{\partial x}dx + \frac{\partial u} {\partial y}dy$
If you divide by du,
$\displaystyle \frac{\mathrm{d} u}{\mathrm{d}u }= \frac{\partial u}{\partial x}\frac{\mathrm{d} x}{\mathrm{d}u } + \frac{\partial u} {\partial y}\frac{\mathrm{d} y}{\mathrm{d}u }$
Which is just another relation between differentials. Just like a linear equation
z=Ax+By
can be divided by z to give:
z/z=Ax/z + By/z.
When in doubt, write the independent variables the dependent variables are a function of.
$\displaystyle du(x,y)=\frac{\partial u(x,y)}{\partial x}dx + \frac{\partial u(x,y)} {\partial y}dy$
You can't do it for the OP.

$\displaystyle \partial u$ and $\displaystyle \partial x$ are not differentials, so there is no cancellation.

Last edited by zylo; January 11th, 2017 at 07:04 PM. Reason: typ

January 11th, 2017, 08:40 PM   #25
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Quote:
 Originally Posted by v8archie Since $\partial d u$ and $\partial d x$ do not feature in your post, so the fact that they aren't differentials is irrelevant except that it casts serious doubt on the idea that you know what you are talking about.
$\displaystyle \partial d u$ What is that?

However, strangely enough,
$\displaystyle \frac{\partial du}{\partial dx}$ makes sense because an equation in differentials,

$\displaystyle du=u_{x}dx+u_{y}dy$
is a linear relation between the variables du, dx and dy.

So if du=z, dx=v, and dy=w, then
$\displaystyle z=u_{x}v+u_{y}w$ and
$\displaystyle \frac{\partial du}{\partial dx}= \frac{\partial z}{\partial v} = u_{x}$

Last edited by zylo; January 11th, 2017 at 08:46 PM.

 January 11th, 2017, 08:47 PM #26 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,439 Thanks: 2114 Math Focus: Mainly analysis and algebra It was supposed to be $\partial u$ and $\partial x$ - they don't exist.
January 11th, 2017, 09:17 PM   #27
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Quote:
 Originally Posted by v8archie It was supposed to be $\partial u$ and $\partial x$ - they don't exist.
Of course not and nobody said they did. But $\displaystyle \frac{\partial u(x,y,z)}{\partial x}$ does exist.

Nevertheless your typo led to an interesting point- see my previous post.

But what is the point of alleging something doesn't exist that nobody said existed.

January 12th, 2017, 03:37 PM   #28
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Quote:
 Originally Posted by zylo However, strangely enough,
I'm pretty certain that you have written more nonsense.

And you were the first to talk about $\partial u$ and $\partial x$.

But, as you may have guessed fron the way I deleted a post above before you had replied to it, I'm getting tired of this subject.

 January 15th, 2017, 03:03 PM #29 Newbie   Joined: Jan 2017 From: Canada Posts: 6 Thanks: 1 Thanks to everyone for taking the time to respond to this post. It seems to have ignited a fiery discussion! To answer the question about "What are x,y,and z functions of?" the following should help: u = u(x, y, z) x = x(u, v, w) v = v(x, y, z) y = y(u, v, w) w = w(x, y, z) z = z(u, v, w) So x, y, z, u, v, and w are very much overloaded symbols that represent: 1) the axis of coordinate systems, 2) the value along the coordinates, and 3) a function defining that value. x, y, z represent a Cartesian coordinate set and u, v, w represent another, possible curvilinear non-orthogonal, coordinate set; the two are related by the six functions given above. At the risk of using jargon without learning basic principles these are all a lead-in to defining the covariant and contravariant base vectors of a curvilinear coordinate set. I wasn't looking for the definition of partial derivatives, that I have, I was hoping for some inside help to solving that original equality ... and the answer is you can't, "you have to have come across that type of equation before"! For details on the introduction to tensors that this is leading to, and for your reading pleasure, refer to the bottom of page 17 and page 18 of: https://www.grc.nasa.gov/www/k-12/Nu...2002211716.pdf Last edited by photogo01; January 15th, 2017 at 03:06 PM.
January 15th, 2017, 03:57 PM   #30
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Also, as a follow up ...
Quote:
 Originally Posted by zylo Definition: $\displaystyle \frac{\partial u}{\partial x}=Lim\frac{u(x+\Delta x,y,z) - u(x,y,z)}{\Delta x}$
is the definition of a partial derivative not of a differential.

Examples of differentials are ds, dx, and dy which can be viewed as infinitesimal lengths. I erred in seeing âˆ‚x as dx and thinking "differentials".

A great introduction to differentials starts on page 93 of:
https://ocw.mit.edu/ans7870/resource...s/Calculus.pdf

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