Question about the differential in Calculus

[agapito]

Question about the differential in Calculus.
Assume a function y = f(x) , differentiable everywhere. Now we have for some Δx
Δy = f(x + Δx) - f(x)

The differential of x, is defined as “dx”, can be any real number, and dx = Δx

The differential of y, is defined by “dy” and
dy = f’(x) dx

Clearly,
Δy ≈ dy, depending on the magnitude of Δx.
In calculus an expression like “dx” usually denotes something infinitesimally small.
Why is it necessary to have dy and dx used as real numbers of some magnitude? In specifying and solving calculus problems are not the usual symbols sufficient?
Is it just a matter of notational convenience?

 

[I like Serena]

Hi agapito,

$\Delta x$ is any real number.
However, $dx$ is not "any real number", and it is also not a "real number of some magnitude".
Where did you get that?

Instead it is an "infinitesimally" small real number, and as such not a real number at all.
It is $\Delta x$ taken to the limit where it approaches $0$.
Put otherwise, on the real line $dx$ is smaller than any non-zero real number.
It means for instance that $\frac 1{dx}$ is greater than any real number (an infinity), and as such not real number either.

When we write $f'(x)=\frac{dy}{dx}$, this actually means $f'(x)=\lim\limits_{\Delta x\to 0} \frac{\Delta y}{\Delta x}$, where $\Delta y=f(x+\Delta x)-f(x)$.

 

[agapito]

Many thanks for your reply, Klaas.

$\Delta x$ is any real number.
However, $dx$ is not "any real number", and it is also not a "real number of some magnitude".
Where did you get that?

From several calculus books. The existence of a separate "infinitesimal number" system is acknowledged but generally not recommended as it intruduces many complications.

By definition of dy,

dy = f'(x) dx, where dx = Δx

The linkage between the definition for "dx" and "dy", as real numbers of any magnitude, is the expression:

(lim Δx-->0) dy = Δy, which follows from the continuity of f(x).

The above is what all books I consulted present. Having learned calculus (many years ago!) where "dx" means something infinitesimal, this treatment was confusing, the reason for my question. Do you think it worthwhile to attempt to learn about infinitesimals?

Thanks again for your interest.

 

[topsquark]

Many thanks for your reply, Klaas.

$\Delta x$ is any real number.
However, $dx$ is not "any real number", and it is also not a "real number of some magnitude".
Where did you get that?

From several calculus books. The existence of a separate "infinitesimal number" system is acknowledged but generally not recommended as it intruduces many complications.

By definition of dy,

dy = f'(x) dx, where dx = Δx

The linkage between the definition for "dx" and "dy", as real numbers of any magnitude, is the expression:

(lim Δx-->0) dy = Δy, which follows from the continuity of f(x).

The above is what all books I consulted present. Having learned calculus (many years ago!) where "dx" means something infinitesimal, this treatment was confusing, the reason for my question. Do you think it worthwhile to attempt to learn about infinitesimals?

Thanks again for your interest.

One way to define dx is to take smaller and smaller \(\displaystyle \Delta x\)'s. \(\displaystyle \Delta x\) has a size... dx does not.

Now, where you will see a difference between what you are saying is to compare Experimental Physics and Mathematics. When we do an experiment then we often take \(\displaystyle dx \to \Delta x\) in order to do numerical calculations. But recognize that this introduces errors into the system.

IMHO: The infinitesimals have great value and should be understood at some point.. but usually that point is after you've studied for several years. For now I'd recommend recognizing that there is a big difference but when you think of it think of dx as a very very small \(\displaystyle \Delta x\).

-Dan

 

[I like Serena]

While I'm at it...

Infinitesimals are part of the so called Hyperreal numbers.
And indeed, they come with trouble, since they break down with various operations like division and subtraction.

For engineers it doesn't matter much, since they will just treat infinitesimals as "small real numbers", which works well enough in most cases.
It's mostly mathematicians that don't really accept that point of view. And there are indeed edge cases where things go wrong.

But there is yet another definition for "differentials". Let's not call them infinitesimals.
In differential geometry, a differential or push forward is defined as a function that maps a vector.
More specifically, we can choose $dx$ as a function that maps $(1,0)$ to $1$, and that maps $(0,1)$ to $0$.
In the same way, we can pick $dy$ as a function that maps $(1,0)$ to $0$, and $(0,1)$ to $1$.

Applied to your example of $dy=f'(x)dx$, the vector $(1,f'(x))$ is the tangential vector to the curve $(x,f(x))$.
The differential $dx$ is a function that maps $(1,f'(x))$ to $1$, and the differential $dy$ is a function that maps it to $f'(x)$.
Consequently we have that $dy(v) = f'(x)dx(v)$ for any vector $v$ at the point $(x,f(x))$, or $dy=f'(x)dx$ for short.

The beauty is that with this definition of differentials, the mathematical framework is valid, robust, and in particular applies to more general manifolds, like space-time.

 

[agapito]

One way to define dx is to take smaller and smaller \(\displaystyle \Delta x\)'s. \(\displaystyle \Delta x\) has a size... dx does not.

Now, where you will see a difference between what you are saying is to compare Experimental Physics and Mathematics. When we do an experiment then we often take \(\displaystyle dx \to \Delta x\) in order to do numerical calculations. But recognize that this introduces errors into the system.

IMHO: The infinitesimals have great value and should be understood at some point.. but usually that point is after you've studied for several years. For now I'd recommend recognizing that there is a big difference but when you think of it think of dx as a very very small \(\displaystyle \Delta x\).

-Dan

Thanks for your guidance, Dan.

 

[agapito]

While I'm at it...

Infinitesimals are part of the so called Hyperreal numbers.
And indeed, they come with trouble, since they break down with various operations like division and subtraction.

For engineers it doesn't matter much, since they will just treat infinitesimals as "small real numbers", which works well enough in most cases.
It's mostly mathematicians that don't really accept that point of view. And there are indeed edge cases where things go wrong.

But there is yet another definition for "differentials". Let's not call them infinitesimals.
In differential geometry, a differential or push forward is defined as a function that maps a vector.
More specifically, we can choose $dx$ as a function that maps $(1,0)$ to $1$, and that maps $(0,1)$ to $0$.
In the same way, we can pick $dy$ as a function that maps $(1,0)$ to $0$, and $(0,1)$ to $1$.

Applied to your example of $dy=f'(x)dx$, the vector $(1,f'(x))$ is the tangential vector to the curve $(x,f(x))$.
The differential $dx$ is a function that maps $(1,f'(x))$ to $1$, and the differential $dy$ is a function that maps it to $f'(x)$.
Consequently we have that $dy(v) = f'(x)dx(v)$ for any vector $v$ at the point $(x,f(x))$, or $dy=f'(x)dx$ for short.

The beauty is that with this definition of differentials, the mathematical framework is valid, robust, and in particular applies to more general manifolds, like space-time.

Thanks. Unfortunately I cannot read your reply due to format of algebraic expressions. Is there any way to correct this so they display in "normal" format?

 

[I like Serena]

Thanks. Unfortunately I cannot read your reply due to format of algebraic expressions. Is there any way to correct this so they display in "normal" format?

What do you mean?
Is there a problem rendering formulas on the device you're using?

 

[agapito]

While I'm at it...

Infinitesimals are part of the so called Hyperreal numbers.
And indeed, they come with trouble, since they break down with various operations like division and subtraction.

For engineers it doesn't matter much, since they will just treat infinitesimals as "small real numbers", which works well enough in most cases.
It's mostly mathematicians that don't really accept that point of view. And there are indeed edge cases where things go wrong.

But there is yet another definition for "differentials". Let's not call them infinitesimals.
In differential geometry, a differential or push forward is defined as a function that maps a vector.
More specifically, we can choose $dx$ as a function that maps $(1,0)$ to $1$, and that maps $(0,1)$ to $0$.
In the same way, we can pick $dy$ as a function that maps $(1,0)$ to $0$, and $(0,1)$ to $1$.

Applied to your example of $dy=f'(x)dx$, the vector $(1,f'(x))$ is the tangential vector to the curve $(x,f(x))$.
The differential $dx$ is a function that maps $(1,f'(x))$ to $1$, and the differential $dy$ is a function that maps it to $f'(x)$.
Consequently we have that $dy(v) = f'(x)dx(v)$ for any vector $v$ at the point $(x,f(x))$, or $dy=f'(x)dx$ for short.

The beauty is that with this definition of differentials, the mathematical framework is valid, robust, and in particular applies to more general manifolds, like space-time.

Thanks. Unfortunately I cannot read your reply due to format of algebraic expressions. Is there any way to correct this so they display in "normal" format?

What do you mean?
Is there a problem rendering formulas on the device you're using?

Yes, They are read as "$(1,f'(x))$ to $1$", for example. I'm reading them in my Windows computer. Is there some way around this? Thanks

 

[HOI]

Many thanks for your reply, Klaas.

$\Delta x$ is any real number.
However, $dx$ is not "any real number", and it is also not a "real number of some magnitude".
Where did you get that?

From several calculus books. The existence of a separate "infinitesimal number" system is acknowledged but generally not recommended as it intruduces many complications.

By definition of dy,

dy = f'(x) dx, where dx = Δx

The linkage between the definition for "dx" and "dy", as real numbers of any magnitude, is the expression:

(lim Δx-->0) dy = Δy, which follows from the continuity of f(x).

The above is what all books I consulted present. Having learned calculus (many years ago!) where "dx" means something infinitesimal, this treatment was confusing, the reason for my question. Do you think it worthwhile to attempt to learn about infinitesimals?

Thanks again for your interest.

I have seen many Calulus and analysis texts and NONE of them have said that! I suspect you have misunderstood. "Real Analysis" by George Thomas (old but my favorite) defines dy= f'(x)dx and does NOT define dx as equal $\Delta x$ but just a s symbol representing an infinitesmal change in x. Both dy and dx are purely symbolic and cannot appear alone, either in an equation together or in an integral.