# B Is this proper or it is abuse of notation?

#### fbs7

Say

$z = f(x,y)$

then I learned that

$dz = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$
$dz = f_x*dx + f_y*dy$

The question is whether this expression on $dz$ is really proper. The question comes from this: I know the definition of say $\frac{dz}{dt}$ as

$\frac{dz}{dt}=lim_{\Delta t->0}\frac{z(t+\Delta t)-z(t)}{\Delta t}$

I understand $dz/dt$ as a ratio, and I can live with extending the notation a little bit to define a thingie $dz = (something) dt$, but in my mind that only makes sense as I understand that's a shorthand for the ratio $dz/dt$.

But how about the expression of $dz$ in terms of $dx$ and $dy$? I know it works fine and is extremely practical (I use it to solve differential equations), and it kinda makes sense as I know I can bring it back to a ratio being a sum of two ratios, but that seems suspiciously like an abuse of notation, as it is it implies that the thingies $dx$ and $dy$ exist by themselves and form an algebra such that I can go around adding, subtracing, dividing and multiplying.

Well, I never heard of that algebra, but if there's truly an "algebra of infinitesimals", then I could write things like

$du=dx*e^{dy/dz-dw/dt}+\sqrt{dx*dy+dw*dt}$

what sounds preposterous to me (maybe it isn't). So, is there indeed an algebra on infinitesimals $dx, dy, dz,...$ or is that just a practical shorthand that only makes sense if you restrict it to linear combination forms?

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#### Math_QED

Homework Helper
In your context it is probably practical shorthand. It can be made rigorous. For example in differential geometry.

#### Mark44

Mentor
Say

$z = f(x,y)$

then I learned that

$dz = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$
$dz = f_x*dx + f_y*dy$

The question is whether this expression on $dz$ is really proper. The question comes from this: I know the definition of say $\frac{dz}{dt}$ as

$\frac{dz}{dt}=lim_{\Delta t->0}\frac{z(t+\Delta t)-z(t)}{\Delta t}$
$\frac{dz}{dt}$ makes sense only if both x and y are functions of t alone. If x and y don't depend on t, then the differential of z (or total differential of z) can be written with either of the notations you have above.
fbs7 said:
I understand $dz/dt$ as a ratio, and I can live with extending the notation a little bit to define a thingie $dz = (something) dt$, but in my mind that only makes sense as I understand that's a shorthand for the ratio $dz/dt$.

But how about the expression of $dz$ in terms of $dx$ and $dy$?
Again, if x and y aren't functions of t, then dz is necessarily in terms of the two partials.
fbs7 said:
I know it works fine and is extremely practical (I use it to solve differential equations), and it kinda makes sense as I know I can bring it back to a ratio being a sum of two ratios, but that seems suspiciously like an abuse of notation, as it is it implies that the thingies $dx$ and $dy$ exist by themselves and form an algebra such that I can go around adding, subtracing, dividing and multiplying.

Well, I never heard of that algebra, but if there's truly an "algebra of infinitesimals", then I could write things like

$du=dx*e^{dy/dz-dw/dt}+\sqrt{dx*dy+dw*dt}$
I have no idea what you are trying to do here.
fbs7 said:
what sounds preposterous to me (maybe it isn't). So, is there indeed an algebra on infinitesimals $dx, dy, dz,...$ or is that just a practical shorthand that only makes sense if you restrict it to linear combination forms?

#### fresh_42

Mentor
2018 Award
There are a few things going on here. In general we consider the tangent plane of $f(x,y)$ at a certain point, which isn't noted here. However, we have a two dimensional tangent space and a vector in this tangent plane, if we regard it as a linear space with the origin at the point where the tangent plane touches $f(x,y)$.
This is given as a linear combination of two basis vectors: $f(\vec{p}) = c_x \vec{b_x} + c_y \vec{b_y}$.

What I now will say might earn me a wild debate about covariance, cotangents and differential forms, but I'll make the simplification anyway.

The formula $\vec{p} = c_x \vec{b_x} + c_y \vec{b_y}$ is in principal the same as $dz = \dfrac{\partial f}{\partial x} dx + \dfrac{\partial f}{\partial y}dy$ with coefficients as the values of the partial derivatives at this point and the differentials $dx,dy$ as the vectors which span the tangent space. This is not 100% accurate$(^*)$, as the differentials are linear functions themselves and not geometric basis vectors, but it describes what is going on.

Now we have $\dfrac{dz}{dt}$ which is actually a geometric tangent vector, which points in the direction of a path on $f(x,y)$ which is parameterized by $t$. I.e. we have no specified direction here. The direction is given by a path along which we build the limit. So this is coordinate free, no basis vectors given. If you write this path as $t \longmapsto \gamma(t) = (x(\gamma(t)), y(\gamma(t))$ in $x,y$ coordinates, then you end up with the same formula for $dz$.

$(^*)$ (All variables here are vectors, so $x$ actually means $(x,y)$ here.)
One of the difficulties with differentials and derivatives is to figure out what is what. We generally have in the Weierstraß formulation $$\mathbf{f(x_{0}+v)=f(x_{0})+(D_{x_0}(f))(v)+r(v)}$$
That is in words $f(x_0+v)$ at point $p=x_0$ varied in direction $v$ a little bit $(v \to 0)$ can be written as a start at $p=x_0$ with value $f(x_0)$ plus a straight line at this point, the tangent $D_{x_0}$ at this point on $f$ in direction $v$ plus an error $r(v)$ we make, which runs faster to zero than linear, if $v$ goes to $0$.
If we rewrite the formula as $\left( D_{x_0}(f) \right)\left( \dfrac{v}{||v||} \right) = \dfrac{f(x_0+v)-f(x_0)}{||v||} - r\left( \dfrac{v}{||v||} \right)$ we get the usual differentiation limit.
The essential part is the interpretation of the derivative $D_{x_0}(f)(v)$. Do we mean the function
• $v \longmapsto D_{x_0}(f)(v)$
• $x_0 \longmapsto \left( v \longmapsto D_{x_0}(f)(v) \right)$
• $f \longmapsto f\,'=D(f)$
• $f \longmapsto (x_0 \longmapsto D_{x_0}(f))$
or simply the slope of the tangent, which is the real number $\left( D_{x_0}(f) \right)\left( \dfrac{v}{||v|||} \right)\,?$ You see, people say derivative and mean any of those; which one depends on the context. The last two are the limit formulation for $\dfrac{dz}{dt}$, the first is the total differential $dz$, and the middle ones what is normally meant at school. But all are the same thing, only under different perspectives.

#### WWGD

Gold Member
dz here is the total differential of z,the differential of a function of two or more variables with respect to a single parameter. This is the local change of the function along the tangent line, plane, etc.

#### fbs7

I have no idea what you are trying to do here.
Well, the question is about having formulas involving $dx, dy, dz$ and so forth terms. I know this formula makes sense:

$\frac{dy}{dx}=f(x)$

then this becomes this formula, which seems to make sense too, because it's just the same derivative just rewritten a bit:

$dy=f(x)dx$

And I know that this makes sense:

$\frac{df(x,y)}{dt}=f_x(x,y)*\frac{dx}{dt}+f_y(x,y)*\frac{dy}{dt}$

which somehow becomes this

$df(x,y)=f_x(x,y)*dx+f_y(x,y)*dy$

Now the right side seems more complicated, and is looking much more like a regular algebraic expression, in this format:

$df=g( x, y, dx, dy )=f_x(x,y)*dx+f_y(x,y)*dy$

It seems that we're treating $df, dx, dy$ as regular variables, just like x and y. If the $g( x, y, dx, dy )$ above is indeed a well-formed function, then one can suppose you make just make any other well formed function with $dx$ and $dy$ and that should make sense too, like

$df = sin(dx^2)+tan(dy*e^{dx})$

but this looks completely wacky too me. Assuming it's wacky, then what's in the nature of the formula $g(x,y,dx,dy)$ that makes

$df=f_x(x,y)*dx+f_y(x,y)*dy$ ==> acceptable
$df = sin(dx^2)+tan(dy*e^{dx})$ ==> out of whack

?

#### WWGD

Gold Member
Well, the question is about having formulas involving $dx, dy, dz$ and so forth terms. I know this formula makes sense:

$\frac{dy}{dx}=f(x)$

then this becomes this formula, which seems to make sense too, because it's just the same derivative just rewritten a bit:

$dy=f(x)dx$

And I know that this makes sense:

$\frac{df(x,y)}{dt}=f_x(x,y)*\frac{dx} {dt}+f_y(x,y)*\frac{dy}{dt}$

which somehow becomes this

$df(x,y)=f_x(x,y)*dx+f_y(x,y)*dy$

Now the right side seems more complicated, and is looking much more like a regular algebraic expression, in this format:

$df=g( x, y, dx, dy )=f_x(x,y)*dx+f_y(x,y)*dy$

It seems that we're treating $df, dx, dy$ as regular variables, just like x and y. If the $g( x, y, dx, dy )$ above is indeed a well-formed function, then one can suppose you make just make any other well formed function with $dx$ and $dy$ and that should make sense too, like

$df = sin(dx^2)+tan(dy*e^{dx})$

but this looks completely wacky too me. Assuming it's wacky, then what's in the nature of the formula $g(x,y,dx,dy)$ that makes

$df=f_x(x,y)*dx+f_y(x,y)*dy$ ==> acceptable
$df = sin(dx^2)+tan(dy*e^{dx})$ ==> out of whack

?
But the total differential you use assumes dependence on a single parameter for one , starting from a general form f(x(t),y(t),....) which does not seem to match your format.

#### fbs7

Right, so, asking another way... I know this is a real, well-formed formula (ie, it is a function too):

$f(x(t),y(t),t) = f_x(x(t),y(t),t)*\frac{dx(t)}{dt}+f_y(x(t),y(t),t)*\frac{dy(t)}{dt}$

And I see this being written as an acceptable formula:

$df(x(t),y(t),t) = f_x(x(t),y(t),t)*dx(t) + f_y(x(t),y(t),t)*dy(t)$

If that's an acceptable formula, then I should be able to define a function with that formula, so the right side should be a function too, same way as the first one. Is that a real function?

Now, if that's a well formed formula and a real function, then what stops me from writing other well formed (but crazy-looking) formulas and call them a function of some sort, as

$df(x(t),y(t),t)$ = whacky-expression-with-dx-and-dy

#### WWGD

Gold Member
Right, so, asking another way... I know this is a real, well-formed formula (ie, it is a function too):

$f(x(t),y(t),t) = f_x(x(t),y(t),t)*\frac{dx(t)}{dt}+f_y(x(t),y(t),t)*\frac{dy(t)}{dt}$

And I see this being written as an acceptable formula:

$df(x(t),y(t),t) = f_x(x(t),y(t),t)*dx(t) + f_y(x(t),y(t),t)*dy(t)$

If that's an acceptable formula, then I should be able to define a function with that formula, so the right side should be a function too, same way as the first one. Is that a real function?

Now, if that's a well formed formula and a real function, then what stops me from writing other well formed (but crazy-looking) formulas and call them a function of some sort, as

$df(x(t),y(t),t)$ = whacky-expression-with-dx-and-dy
If I understood your notation, you have three variable s depending on t, the third one being z(t)=t. So you need to expand accordingly.

#### WWGD

Gold Member
In order to operate , d is defined only on inputs of the form $(x_1(t), x_2(t),.. ,x_n(t))$.

#### fbs7

In order to operate , d is defined only on inputs of the form $(x_1(t), x_2(t),.. ,x_n(t))$.
Hmm.. I see... so there's no such thing as f( dx(t) ), correct?

That is, d is an operator, and x(t) is the variable... dx(t) is not a variable, correct?

That is, df(t) is neither a function or a variable.. it's an operator on a function?

#### WWGD

Gold Member
Hmm.. I see... so there's no such thing as f( dx(t) ), correct?

That is, d is an operator, and x(t) is the variable... dx(t) is not a variable, correct?
Unless you define a function dx(t) of the variable t, no , it is not.

#### fbs7

Wow... a whole new world is opened!!! I gotta figure out this "operator" kind of thingie!! It's neither a function nor a variable, it's something else... very mysterious!

Gotta read about it! I thought there were only numbers (an axiomatic thingie), variables (something that can have a value of a number) and functions (associations between numbers or variables), so I've been using dx, dy, dz as regular variables, when they are actually something different - that's exciting! A world opens! Thank you!

#### WWGD

Gold Member
Basically you have a function in which the arguments are functions, though the functions themselves depend on an input variable.

#### Mark44

Mentor
Well, the question is about having formulas involving $dx, dy, dz$ and so forth terms. I know this formula makes sense:

$\frac{dy}{dx}=f(x)$
No, it doesn't, not if you are given that y = f(x).
fbs7 said:
then this becomes this formula, which seems to make sense too, because it's just the same derivative just rewritten a bit:

$dy=f(x)dx$
No, if y = f(x), then dy = f'(x)dx. Note that on the right side there is a derivative, something that you omitted in both equations.
fbs7 said:
And I know that this makes sense:

$\frac{df(x,y)}{dt}=f_x(x,y)*\frac{dx}{dt}+f_y(x,y)*\frac{dy}{dt}$

which somehow becomes this

$df(x,y)=f_x(x,y)*dx+f_y(x,y)*dy$
A nonrigorous explanation is that you're just multiplying both sides by dt. And again, this isn't valid unless x and y are functions of t.
fbs7 said:
Now the right side seems more complicated, and is looking much more like a regular algebraic expression, in this format:

$df=g( x, y, dx, dy )=f_x(x,y)*dx+f_y(x,y)*dy$

It seems that we're treating $df, dx, dy$ as regular variables, just like x and y.
df and dy are differentials, where df is defined in terms the two partials of f and dx and dy. If y is a function of x, then dy = g'(x)dx.
fbs7 said:
If the $g( x, y, dx, dy )$ above is indeed a well-formed function, then one can suppose you make just make any other well formed function with $dx$ and $dy$ and that should make sense too, like

$df = sin(dx^2)+tan(dy*e^{dx})$
No. I can't think of any scenarios where this is meaningful.
fbs7 said:
but this looks completely wacky too me. Assuming it's wacky, then what's in the nature of the formula $g(x,y,dx,dy)$ that makes

$df=f_x(x,y)*dx+f_y(x,y)*dy$ ==> acceptable
$df = sin(dx^2)+tan(dy*e^{dx})$ ==> out of whack

#### fresh_42

Mentor
2018 Award
Hmm... trying to rephrase the question. I think basis of my question is this: are these variables?

$dx(t), dy(t)$

if they are variables, can I write a function on them, like

$f(dx(t),dy(t)) = f_1(t)*dx(t) + f_2(t)*dy(t)$

Is that a true function?
Here is an exemplary calculation how differential forms are used:

You could start to read the Wiki pages on the subject:
https://en.wikipedia.org/wiki/Differential_of_a_function
https://en.wikipedia.org/wiki/Differential_form

As mentioned earlier, the situation is a bit complex and requires precise handling. If you want to have a list what differentiation can mean, cp. https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/ (section 1), and I didn't even use the word slope. It is extremely important to clear the context first, will say you cannot mix the various meanings to create inconsistencies, which are finally based on an improper start.

#### fbs7

dz=fx(x,y)dx+fy(x,y)dydz=fx(x,y)dx+fy(x,y)dy

A nonrigorous explanation is that you're just multiplying both sides by dt. And again, this isn't valid unless x and y are functions of t.
I see, so that form as is, it is indeed abusive? That seems to be quite frequent abuse. For example, at Wikipedia

https://en.wikipedia.org/wiki/Differential_of_a_function#Differentials_in_several_variables

he uses exactly that format on the section "Differentials in several variables", and explicitly calls x and y independent variables. When I first saw that expression I thought it was fishy, as I understood a derivative as being a ratio... so where's the ratio on dx and dy if they are independent?... very confusing concept!

For comparison, this makes sense in my mind:

$df(x(t),y(t))dt=\frac{∂f(x(t),y(t))}{∂x(t)}.\frac{dx(t)}{dt}+\frac{∂f(x(t),y(t))}{∂y(t)}.\frac{dy(t)}{dt}$

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#### Mark44

Mentor
I didn't say it was "abusive," only that it was not a rigorous explanation. These aren't synonyms.

When you're first taught about the derivative, it is typically explained as being a ratio, but a bit later, most textbooks provide an explanation of how differentials are defined.

#### fbs7

Here is an exemplary calculation how differential forms are used:

You could start to read the Wiki pages on the subject:
https://en.wikipedia.org/wiki/Differential_of_a_function
https://en.wikipedia.org/wiki/Differential_form

As mentioned earlier, the situation is a bit complex and requires precise handling. If you want to have a list what differentiation can mean, cp. https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/ (section 1), and I didn't even use the word slope. It is extremely important to clear the context first, will say you cannot mix the various meanings to create inconsistencies, which are finally based on an improper start.
Holy choo-choo! This is a brilliant link: https://en.wikipedia.org/wiki/Differential_of_a_function !

It's like a lot of pieces suddenly fall in place. So $df(x)$ is really defined as

$df(x)=df(x,Δx)=f′(x).Δx$

what means $dx=dx(x,Δx)=Δx$

This way there's no need to any jubba-jubba or jibby-jabby around... Now it's so simple, $df(x)$ is really the definition of a function on $x$ and $Δx$ and it exists on its own, without needing anything else! Extraordinary! The guys that come up with these definitions are really sensational!!

I gotta change my books, I reckon; my books focus in the ratio point of view, and this one is so much simpler.

From that point of view, this is then a fine expression (maybe a definition too?), it is a function even when x and y are independent:

$df(x,y)=f_x(x,y)*dx+f_y(x,y)*dy = f_x(x,y).\Delta x + f_y(x,y).\Delta y$

Not if I can figure out what an "operator" is...hmm... where's my link to Wikipedia...

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#### Mark44

Mentor
Holy choo-choo! This is a brilliant link: https://en.wikipedia.org/wiki/Differential_of_a_function !

It's like a lot of pieces suddenly fall in place. So $df(x)$ is really defined as

$df(x)=df(x,Δx)=f′(x).Δx$
No. If you look more carefully, you'll see that it's defined this way:
$df(x), \Delta x) = f'(x)\Delta x$,
but df(x) is defined as $f'(x)dx$.

dx and $\Delta x$ are not the same.
fbs7 said:
what means $dx=dx(x,Δx)=Δx$

This way there's no need to any jubba-jubba or jibby-jabby around... Now it's so simple, $df(x)$ is really the definition of a function on $x$ and $Δx$ and it exists on its own, without needing anything else!
Again, no. See above.
fbs7 said:
Extraordinary! The guys that come up with these definitions are really sensational!!

I gotta change my books, I reckon; my books focus in the ratio point of view, and this one is so much simpler.

From that point of view, this is then a fine expression (maybe a definition too?), it is a function even when x and y are independent:

$df(x,y)=f_x(x,y)*dx+f_y(x,y)*dy = f_x(x,y).\Delta x + f_y(x,y).\Delta y$
No. This part is correct: $df(x,y)=f_x(x,y)*dx+f_y(x,y)*dy$, but it's not equal to the last expression. An important quality in being successful in mathematics is attention to detail...
fbs7 said:
Not if I can figure out what an "operator" is...hmm... where's my link to Wikipedia...

"Is this proper or it is abuse of notation?"

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