Definition of derivative - infinitesimal approach, help :)

[christian0710]

Hi I'm reading Elementary calculus - an infinitesimal approach and just wan't to make sure my understanding of what dy, f'(x) and dx means is correct.

I do understand the basic idea: You make the secant between 2 points on a graph approach one of the points and at this point you get the tangent to the graph which is the derivative of f at that point and tells you the slope of the function at that point. But I want the correct mathematical understanding of it.

The book argues that

Δy = the change in y along a curve between 2 points (I assume it's a secant)
dy = change in y along the tangent line to that curve between 2 points (The differential)

Then it defines that

Δy = f(x+Δx)-f(x)
dy = f'(x)Δx

and mentiones that
"let y=f(x). Suppose f'(x) exists at a cetain point x, and Δx is the infinitesimal, then Δy is infinitesimal and"

Δy = f'(x)Δx+ εΔx

and prooves that

Δy/Δx ≈ f'(x)
Δy/Δx = f'(x) + ε
Δy = f'(x)Δx + εΔx

My first question is This:

Is f(x+Δx)-f(x) and f'(x)Δx+ εΔx Equal? If they both are equal to Δy then i assume they are?

Is this the correct understanding:
This expression Δy/Δx only approaches f'(x): Δy/Δx = (y2-y1)/(x2-x1) = (f'(x)Δx+εΔx)/((x+εΔx-x)) ≈ f'(x)
While this expression is equal dy/dx = f'(x)

The difference between
dy/dx = f'(x)
and
Δy/Δx ≈ f'(x)

is that Δy= dy+ εΔx contains that extra "εΔx" term and therefore is bigger than dy and is also the secant between 2 points along the graph, whereas dy is an infinitesimal small movement between 2 points on the tangent to the graph. Because dy=f'(x)*Δx equals the term f'(x)*Δx this tells us that dy is an infinitesimally small change in y between the point at the tangent and another point on the tangent infinitesimally close to that point. The Δx in f'(x)Δx shows us that it's a "change in y" corresponding to a infinitesimally small change in x so (x2-x1), hence Δx?

If the f'(x) exists then the differential dy and the increment Δy MUST be infitesimal and so close together that they cannot be seen under the infinitesimal microscope.

So my last question: I guess it's a no-go to treat the symbol dy/dx as a quotient? The book mentions that if dy=f'(x)dx and if dx ≠ 0 then we can rewrite the equation dy/dx = f'(x), is this just lucky that this "trick" works, or is it true that you can treat it like a quotient?

 

[davidmoore63@y]

Is f(x+Δx)-f(x) and f'(x)Δx+ εΔx Equal ...yes Δy/Δx = (y2-y1)/(x2-x1) = (f'(x)Δx+εΔx)/((x+εΔx-x)) ≈ f'(x)
where did you get the denominator x+εΔx-x from? I think it should just be Δx
then the expression reads Δy/Δx = (y2-y1)/(x2-x1) = (f'(x)Δx+εΔx)/(Δx) = f'(x)+ε
and the ε goes to zero as Δx goes to zero.I guess it's a no-go to treat the symbol dy/dx as a quotient?
yes i think it's technically incorrect to consider dy/dx as a quotient. Rather it is the limit of Δy/Δx. If the book says "that if dy=f'(x)dx and if dx ≠ 0 then we can rewrite the equation dy/dx = f'(x)", they are being technically lazy/incorrect. On the other hand casual users of calculus often treat the dy/dx as a quotient because it saves time and we don't want to write the full definition every time we manipulate the equations. But be warned this is just shorthand.

 

[verty]

I guess it's a no-go to treat the symbol dy/dx as a quotient?
yes i think it's technically incorrect to consider dy/dx as a quotient. Rather it is the limit of Δy/Δx.

I have to say, if one wanted to learn infinitesimals to avoid limits, this is certainly bad news. The other bit of bad news is that the secant, when $\Delta{x}$ goes infinitesimal, becomes the tangent. This is the same as being in the limit. If all one is doing is switching the word "infinitesimal" for the word "limit", but still one must use limits or think in terms of being in the limit, I don't see what one has gained.

I haven't read the book in detail, I did look at it about 6 years ago, but I'd be very surprised if $dx$ is allowed to be zero. That seems odd to me. If $dx$ is allowed to be zero, I would shelve this theory and just learn the usual calculus because then it is a load of tosh.

 

[EM_Guy]

So my last question: I guess it's a no-go to treat the symbol dy/dx as a quotient? The book mentions that if dy=f'(x)dx and if dx ≠ 0 then we can rewrite the equation dy/dx = f'(x), is this just lucky that this "trick" works, or is it true that you can treat it like a quotient?

You can treat dy/dx as a quotient. Whether this is technical precise mathematically, I don't know. But for all engineering purposes, you can treat dy/dx as a quotient. In differential equations, you habitually multiply both sides of an equation by dx.

 

[christian0710]

Thank you guys. So when is it okay to treat dy/dx as a quotient? Are there some actual rules or guilde lines? The concept of what the limit is, and what rules apply to dy/dx, and what it TRULY means has always confused me, can someone enlighten me? :)

On the other hand casual users of calculus often treat the dy/dx as a quotient because it saves time and we don't want to write the full definition every time we manipulate the equations.

Just out of curiosity: What is the Full definition?

 

[MidgetDwarf]

Thank you guys. So when is it okay to treat dy/dx as a quotient? Are there some actual rules or guilde lines? The concept of what the limit is, and what rules apply to dy/dx, and what it TRULY means has always confused me, can someone enlighten me? :)

Just out of curiosity: What is the Full definition?

When delta x is not equal to zero. Ussually used for problems of physical importance. For ex. Geometric Shapes (distances is one), physics problems are 2 that come to mind. Leibnezz notation ussually treats the differential as 2 separate things. A quick Google search will give you more information.

 

[verty]

Christian, you have bought so many good calculus books, I suggest you read them instead of this one, for example I think you bought the 3rd edition of Thomas that MidgetDwarf recommended, and possibly you also bought Simmons. I would use them instead of this one because these infinitesimals are only going to make it more confusing.

I'll put down here what I don't like about this book.

1. It calls $\Delta{x}$ an infinitesimal. Perhaps it means that $\Delta{x}$ can be infinitesimal, but for me this is unfortunate and $dx$ should be the infinitesimal. $\Delta{x}$ should be a finite quantity, to have it be both is too confusing. Then it makes sense to say $\Delta{x} \to dx$ (as in, in the limit it gets there).

2. It uses $\epsilon$ as a finite number. But I believe infinitesimal theories usually treat $\epsilon$ as an infinitesimal. It's just too confusing to read a formula where $\Delta{x}$ is infinitesimal but $\epsilon$ is finite.

3. Infinitesimals don't make calculus easier to understand. The usual textbooks are going to have better quality because there are more of them and they are the standard.

Sorry for that, I'll let the thread continue but I wanted to let it be known that IMHO this is not a good way to learn calculus.

 

[micromass]

Hi I'm reading Elementary calculus - an infinitesimal approach and just wan't to make sure my understanding of what dy, f'(x) and dx means is correct.

I do understand the basic idea: You make the secant between 2 points on a graph approach one of the points and at this point you get the tangent to the graph which is the derivative of f at that point and tells you the slope of the function at that point. But I want the correct mathematical understanding of it.

The book argues that

Δy = the change in y along a curve between 2 points (I assume it's a secant)
dy = change in y along the tangent line to that curve between 2 points (The differential)

Then it defines that

Δy = f(x+Δx)-f(x)
dy = f'(x)Δx

and mentiones that
"let y=f(x). Suppose f'(x) exists at a cetain point x, and Δx is the infinitesimal, then Δy is infinitesimal and"

Δy = f'(x)Δx+ εΔx

and prooves that

Δy/Δx ≈ f'(x)
Δy/Δx = f'(x) + ε
Δy = f'(x)Δx + εΔx

My first question is This:

Is f(x+Δx)-f(x) and f'(x)Δx+ εΔx Equal? If they both are equal to Δy then i assume they are?

Yes.

So my last question: I guess it's a no-go to treat the symbol dy/dx as a quotient? The book mentions that if dy=f'(x)dx and if dx ≠ 0 then we can rewrite the equation dy/dx = f'(x), is this just lucky that this "trick" works, or is it true that you can treat it like a quotient?

[/QUOTE]

No, the book is not sloppy, incorrect or full of tosh. It means exactly what is says and it is correct. In the context of the book, $dy/dx$ IS the quotient of $dy$ and $dx$. But notice the difference in definitions, usually $dy/dx$ is defined as the limit of a fraction, this is not the approach here. This is one of the great accomplishments of infinitesimals.

 

[mathman]

Strictly mathematical terms, dy/dx is not a quotient. It is the limit of Δy/Δx as Δx -> 0. It can be safely handled as a quotient as long as you are aware of the limitations.

 

[micromass]

1. It calls $\Delta{x}$ an infinitesimal. Perhaps it means that $\Delta{x}$ can be infinitesimal but for me, this is unfortunate and $dx$ should be the infinitesimal. $\Delta{x}$ should be a finite quantity, to have it be both is too confusing. Then it makes sense to say $\Delta{x} \to dx$ (as in, in the limit it gets there).

Even in standard calculus, this is wrong. The definitions of $\Delta x$ and $dx$ in standard books is exactly the same as the definitions here.

3. Infinitesimals don't make calculus easier to understand. The usual textbooks are going to have better quality because there are more of them and they are the standard.

Saying that there are more books that deal with standard calculus is not really an argument for saying those books are better. Keisler is an excellent book and an excellent way to learn calculus.

 

[MidgetDwarf]

Hi I'm reading Elementary calculus - an infinitesimal approach and just wan't to make sure my understanding of what dy, f'(x) and dx means is correct.

I do understand the basic idea: You make the secant between 2 points on a graph approach one of the points and at this point you get the tangent to the graph which is the derivative of f at that point and tells you the slope of the function at that point. But I want the correct mathematical understanding of it.

The book argues that

Δy = the change in y along a curve between 2 points (I assume it's a secant)
dy = change in y along the tangent line to that curve between 2 points (The differential)

Then it defines that

Δy = f(x+Δx)-f(x)
dy = f'(x)Δx

and mentiones that
"let y=f(x). Suppose f'(x) exists at a cetain point x, and Δx is the infinitesimal, then Δy is infinitesimal and"

Δy = f'(x)Δx+ εΔx

and prooves that

Δy/Δx ≈ f'(x)
Δy/Δx = f'(x) + ε
Δy = f'(x)Δx + εΔx

My first question is This:

Is f(x+Δx)-f(x) and f'(x)Δx+ εΔx Equal? If they both are equal to Δy then i assume they are?

Is this the correct understanding:
This expression Δy/Δx only approaches f'(x): Δy/Δx = (y2-y1)/(x2-x1) = (f'(x)Δx+εΔx)/((x+εΔx-x)) ≈ f'(x)
While this expression is equal dy/dx = f'(x)

The difference between
dy/dx = f'(x)
and
Δy/Δx ≈ f'(x)

is that Δy= dy+ εΔx contains that extra "εΔx" term and therefore is bigger than dy and is also the secant between 2 points along the graph, whereas dy is an infinitesimal small movement between 2 points on the tangent to the graph. Because dy=f'(x)*Δx equals the term f'(x)*Δx this tells us that dy is an infinitesimally small change in y between the point at the tangent and another point on the tangent infinitesimally close to that point. The Δx in f'(x)Δx shows us that it's a "change in y" corresponding to a infinitesimally small change in x so (x2-x1), hence Δx?

If the f'(x) exists then the differential dy and the increment Δy MUST be infitesimal and so close together that they cannot be seen under the infinitesimal microscope.

So my last question: I guess it's a no-go to treat the symbol dy/dx as a quotient? The book mentions that if dy=f'(x)dx and if dx ≠ 0 then we can rewrite the equation dy/dx = f'(x), is this just lucky that this "trick" works, or is it true that you can treat it like a quotient?

No the last part that contains epsilon (delta x). Is considered extremely small so it is an approximation to the curve. As you take the limit In this case, there is no argument that both epsilon and delta x cannot be zero in the last term. As you take the limit of delta x as it approaches infinity. The value is extremely small so it is neglible. So you can ignore it so to speak.

 

[christian0710]

Thank you guys! Now I think i finally understand it 100% Clearly! - the infinitesimal approach to Calculus, so I'll try to explain it:

This equation
Δy =f'(x)*Δx +εΔx

means: an infinitesimal change in y ( an infinitly small distance in y-direction between 2 points ) of a function f can be found as a corresponding change in y of the slope of a tangent to the graph, hence the f'(x)*Δx, PLUS the the infinitemimal small distance between then graph and the tangent, hence then εΔx term.

So this Δy is the vertical y value between 2 infinitely close points on a graph of f.
and this f'(x)*Δx is the y-value between the same 2 points on the tangent to a point on f.
This εΔx tells us that the vertical y-value/distance between 2 points on the graph of f is + εΔx bigger than the vertical distance betwen the 2 points on the tangent to f which is f'(x)Δx

I added two figures to show that εΔx can either be the vertical displacement between the tangent and the graph of f , or the distance between the point (x+Δx, y+Δy) on the tangent and on the grap of f. However in reality εΔx would be MUCH smaller than f'(x)*Δx, but in the figure they look almost the same in size.

Fig 1 where εΔx is the vertical distance between the end point (x+Δx, y+Δy) of the graph of f and the tangent

Fig 2 where εΔx is the vertical distance between the tangent and the graph of f

So if we look at the same equation, and we divide it by Δx,
(Δy =f'(x)*Δx +εΔx)/Δx

we get

Δy/Δx =f'(x) +ε

This equation shows us that the change in Y with respect to a change in X over a infinitesimal distance (between 2 points) only comes near to f'(x), but the actual slope at ONE point f'(x) and the slope between 2 infinitesimal pointsΔy/Δx still separated by a small ε
Δy/Δx ≈ f'(x)

So for the equal signs to hold we need to add ε
Δy/Δx =f'(x) +ε

If we take the standard parts of this ratio, we get

st( Δy/Δx) = f'(x) =dy/dx=(f'(x)*dx)/dx = st(Δy =f'(x)*Δx +εΔx)/Δx

standard parts means that if Δy/Δx is a hyperreal number ( hyperreal numbers includes real numbers + infinitesimals) - consisting of the sum of a real number f'(x) and an infinitsimal ε - then the difference between (f'(x) + ε) -(f'(x)) is infinitesimal, and so the standard part of f'(x) +ε, denoted st(f'(x)+ε) or st(Δy/Δx) is a real number f'(x) which is infinitesimally close to f'(x) +ε .

Δy/Δx = f'(x) +ε
st(Δy/Δx) =st( f'(x) +ε) = f'(x) = dy/dx.

This way of writing it should also be true
Δy/Δx =dy/dx +ε =f'(x) + ε

Regarding the Notation Δy/Δx
If Δy/Δx Is treated as the limit then it's a symbol and not a quotient (Thank you micromass)
If Δy/Δx is treated as a quotient as in the book (which works) then the differential is dy= f'(x)Δx = f'(x)*dx (in the book Δx and dx are the same and both are infinitesimal)End Question: The book successfully treats Δy/Δx as a quotient: Does this work in ANY situation?
Conclusion: If you want to derive a function that shows the slope of the graph of that function with some geometrical intuition, Then you find the slope betrween 2 infinitely close points Δy/Δx , and then by taking the standard parts (by removing everything which is infinitesimal) you have the derivative f'(x) of that function.So i guess that has been the fundamental problem in calculus back in the days: How do you get the instantaneous velocity/slope at one point on a function when you need 2 points to get a slope, well you just invent the infinitesimal numbers, or take the limit and then remove the delta X'es. Smart people :D

 

[Mark44]

I don't understand what you're doing in Figure 2. The line you show is not a tangent to the curve. A line that is tangent to a curve at some point touches the curve at that point.

 

[micromass]

Regarding the Notation Δy/Δx
If Δy/Δx Is treated as the limit then it's a symbol and not a quotient (Thank you micromass)
If Δy/Δx is treated as a quotient as in the book (which works) then the differential is dy= f'(x)Δx = f'(x)*dx (in the book Δx and dx are the same and both are infinitesimal)

I think you mean $dy/dx$ and not $\Delta y/\Delta x$. Even in standard situations, $\Delta y/\Delta x$ is a number. It is $dy/dx$ that is usually defined as a limit, but that here is succesfully defined as a quotient.

End Question: The book successfully treats Δy/Δx as a quotient: Does this work in ANY situation?

Assuming you mean $dy/dx$, yes it always works.

Conclusion: If you want to derive a function that shows the slope of the graph of that function with some geometrical intuition, Then you find the slope betrween 2 infinitely close points Δy/Δx , and then by taking the standard parts (by removing everything which is infinitesimal) you have the derivative f'(x) of that function.So i guess that has been the fundamental problem in calculus back in the days: How do you get the instantaneous velocity/slope at one point on a function when you need 2 points to get a slope, well you just invent the infinitesimal numbers, or take the limit and then remove the delta X'es. Smart people :D

Yes, indeed, that was a very smart solution by some of the smartest people ever to live. But notice that the approach had problems back then. The problem was that nobody really knew what infinitesimals were and what their exact properties are. It is only recently that those questions were answered. Their properties are well-known now and follow mostly from the transfer principle (which is one of the most awesome results out there). This is why the standard approach has been invented: to eliminate any issue with infinitesimals which were seen as badly behaved. So the difference between the standard and the nonstandard approach is that the standard approach is able to formulate everything in function of real numbers, and they do not need extra infinitesimal or infinite numbers.

 

[christian0710]

I don't understand what you're doing in Figure 2. The line you show is not a tangent to the curve. A line that is tangent to a curve at some point touches the curve at that point.

Ahh yes, i see your point: In the book is written "When f'(x) exists then The curve y=f(x) and the tangent line at (x,y) are so close to each other that they cannot be distinguished under an infinitesimal microscope." So i assumed that εΔx could also be interpreted as the distance between the tangent line and the function, but i see now that it makes no sense, so εΔx is to be interprete as figure one. Thank you for clarifying :)

 

[christian0710]

I think you mean dy/dxdy/dx and not Δy/Δx\Delta y/\Delta x. Even in standard situations, Δy/Δx\Delta y/\Delta x is a number. It is dy/dxdy/dx that is usually defined as a limit, but that here is succesfully defined as a quotient.

Yes thank you for pointing that out. dy/dx is what i meant :)

P.s I really like this approach to calculus, it's a great book you recommended back in my other post on calculus book . Thank you :)

 

[christian0710]

So just to be clear with the last part

Δy/Δx Is the slope of a secant between 2 points of the function f
dy/dx is the change in y with respect to x On the tangent line.
dy =f'(x)dx is called the differential and is just a way of rewriting dy/dx = f'(x) and it works,
UPDATE from micromass: so dy and dx are changes in the tangent line and means. the tangent line dy is proportional to the change in the domain dx, and the proportionality factor is f'(x).

just a quick last question:
If dy/dx means a change in y with respect to x, at a specific point, then what exactly does dy=f'(x)*Δx mean? Is it just a notation, or does it have a meaning different from dy/dx?

 

[micromass]

So just to be clear with the last part

Δy/Δx Is the slope of a secant
dy/dx is the change in y with respect to x at One point.
dy =f'(x)dx is called the differential and is just a way of rewriting dy/dx = f'(x) and it works,

just a quick last question:
If dy/dx means a change in y with respect to x, at a specific point, then what exactly does dy=f'(x)*Δx mean? Is it just a notation, or does it have a meaning different from dy/dx?

Do you mean $dy = f^\prime (x) dx$? The $dy$ and $dx$ are changes on the tangent line. It means that the change of the tangent line $dy$ is proportional to the change in the domain $dx$. In this sense, $dy$ and $dx$ are numbers that are related as $dy/dx = f^\prime(x)$ or equivalently as $dy= f^\prime(x) dx$.

 

[Mark44]

So just to be clear with the last part

Δy/Δx Is the slope of a secant
dy/dx is the change in y with respect to x at One point

Also, and more importantly in regard to your figure 1, dy/dx is the slope of the tangent line at the point (x, f(x)).

 

[verty]

To back up what I was saying about the possibility of dx being zero, I quote from here:

Thus differentials (or infinitesimals) dx were ascribed variously the four following properties:

1. dx ≈ 0
2. neither dx = 0 nor dx ≠ 0
3. dx² = 0
4. dx → 0

where “≈” stands for “indistinguishable from”, and “→ 0” stands for “becomes vanishingly small”. Of these properties only the last, in which a differential is considered to be a variable quantity tending to 0, survived the 19th century refounding of the calculus in terms of the limit concept.

This is obviously historical but one can see that originally dx was not allowed to be zero. The points got super duper close but didn't ever land on each other, they were still separated but were infinitely close. And they were close enough that their secant *was* the tangent. That said, this point of view became obsolete when calculus was made rigorous and the limit concept became the standard.

And I must say, I really like the limit concept and would definitely rely on it. I think it is the right notion and one shouldn't need an alternative because limits work perfectly well and are, I believe, easier to understand than infinitesimals.

 

[micromass]

And I must say, I really like the limit concept and would definitely rely on it. I think it is the right notion and one shouldn't need an alternative because limits work perfectly well and are, I believe, easier to understand than infinitesimals.

And if you end up doing things like differential geometry and algebraic geometry, or some parts of physics, then the limit concept proves absolutely inadequate. Infinitesimals give such a huge advantage there, and not understand them well has set me back multiple times. The reality is that differential geometers think in infinitesimals, but write with limits!

 

[christian0710]

Do you mean $dy = f^\prime (x) dx$? The $dy$ and $dx$ are changes on the tangent line. It means that the change of the tangent line $dy$ is proportional to the change in the domain $dx$. In this sense, $dy$ and $dx$ are numbers that are related as $dy/dx = f^\prime(x)$ or equivalently as $dy= f^\prime(x) dx$.

Great, it makes sense! so we do agree on figure 1 that dy=f'(x)dx is the distance in the vertical direction between the 2 points of infinitesimal distance on the tangent line?
And then when we divide dy=f'(x)dx by dx, then this dy/dx=f'(x) is the slope f'(x) at one point on the tangent?

 

[micromass]

Great, it makes sense! so we do agree on figure 1 that dy=f'(x)dx is the distance in the vertical direction between the 2 points of infinitesimal distance on the tangent line?

Well, $dy$ and $dx$ don't need to be infinitesimal, it is defined for every number $dx$ and $dy$. Of course, it will be useful mainly when they're infinitesimal. It means what you said though: it means the change of the tangent line when you change $x$ with $dx$.

And then when we divide dy=f'(x)dx by dx, then this dy/dx=f'(x) is the slope f'(x) at one point on the tangent?

It is the slope of the tangent line, yes.

 

[christian0710]

Since i can't edit my older posts, I'm just going to make a last recap post to summarize everything with corrections amd better a better explanation. Calculus the infinitesimal approach - Thank you so much for the help!

This equation
Δy = f(x +Δx) - f(x)

Is the vertical distance between 2 points (x,y) and (x+Δx,y+Δy) on the graph of f. In our case they are infinitely close togther , so Δy, Δx, ε are infinitesimal.

Δy =f'(x)*Δx + εΔx

This expresses the same as the equation above: Vertical distance between 2 points of infinitesimal distance apart on a graph of f,

However this vertical distance is broken up in 2 parts f'(x)*Δx and εΔx which geometrically can be interpreted like this:

f'(x)*Δx is the the vertical distance between the 2 points (x,y) and (x+Δx , y+f'(x)*Δx) also called ((x+Δx,y+dy) on the tangent to the graph of f, touching f at (x,y)

εΔx is the vertical distance between the end point (x+Δx,y+dy) on the tangent and then end point (x+Δx,y+Δy) on the graph of f.

Δy =f'(x)*Δx + εΔx means: an infinitesimal change in y ( an infinitly small distance in y-direction between 2 points (x,y) and ( (x+Δx,y+Δy) on the graph of f ) of a function f can be found as a corresponding change in y on the tangent to the graph, hence the f'(x)*Δx, PLUS the the infinitemimal small distance between the graph and the tangent between the point (x+Δx,y+Δy) on the graph and ((x+Δx,y+dy) on the tangent, hence the εΔx term.

So if we look at the same equation, and we divide it by Δx,
(Δy =f'(x)*Δx +εΔx)/Δx

we get

Δy/Δx =f'(x) +ε

This equation shows us that the change in Y with respect to a change in X over a infinitesimal distance (between 2 points on f ) only comes near to f'(x), but the actual slope at ONE point f'(x) and the slope between 2 infinitesimal points Δy/Δx still separated by a small ε
Δy/Δx ≈ f'(x)

So for the equal signs to hold we need to add ε
Δy/Δx =f'(x) +ε

If we take the standard parts of this ratio, we get

st( Δy/Δx) = f'(x) =dy/dx=(f'(x)*dx)/dx = st(Δy = (f'(x)*Δx +εΔx)/Δx

standard parts means that if Δy/Δx is a hyperreal number ( hyperreal numbers includes real numbers + infinitesimals) - consisting of the sum of a real number f'(x) and an infinitsimal ε - then the difference between (f'(x) + ε) -(f'(x)) is infinitesimal, and so the standard part of f'(x) +ε, denoted st(f'(x)+ε) or st(Δy/Δx) is a real number f'(x) which is infinitesimally close to f'(x) +ε .

Δy/Δx = f'(x) +ε
st( Δy/Δx) = f'(x) =dy/dx=(f'(x)*dx)/dx = st('(x)*Δx +εΔx)/Δx)

This way of writing it should also be true
Δy/Δx =dy/dx +ε =f'(x) + εΔy = the change in y along a curve between 2 points ( a secant)
dy = is the vertical distance between two points on the tangent line and is equal by multiplying the horizontal distance dx with f′(x)
Δy/Δx Is the slope of a secant between 2 points of the function f.
dy/dx =f'(x) is the change in y with respect to x On the tangent line (So it's the slope of the tangent line) at the point (y,x). and i can be treated as a quotient or as a limit for f as dx aproaches zero.
dy =f'(x)dx So dy and dx are changes on the tangent line, this means the change in y, denoted dy, on the tangent is proportional to the change in x, denoted dx, f'(x) is the proportionality constant/slope at that point.

 

[christian0710]

Well, dy and dx don't need to be infinitesimal, it is defined for every number dx and dy.

Mhh I'm not completely sure how to understand that: I understand that we take the standard parts st(Δy/Δx= st(f'(x) +ε) = dy/dx =f'(x) --> dy =f'(x)dx to eliminate the infinitesimal distance between the tangent and the graph so we only have the change in the tangent.
But am i wrong in asuming that dy means a distance between 2 points on the tangent and this distance is proportional to the distance that we get if we multiply the distance dx by the slope f'(x) as in f'(x) Δx?

What do you mean by dy can be a number? In the figure from the book dy is the distance between the tangent and the graph and that distance is infinitesimal .

Lastly just to make sure: Can you consider dy/dx as a number? because dy/dx = f'(x) and the slope f'(x) is surely a number.

 

[micromass]

Mhh I'm not completely sure how to understand that: I understand that we take the standard parts st(Δy/Δx= st(f'(x) +ε) = dy/dx =f'(x) --> dy =f'(x)dx to eliminate the infinitesimal distance between the tangent and the graph so we only have the change in the tangent.
But am i wrong in asuming that dy means a distance between 2 points on the tangent and this distance is proportional to the distance that we get if we multiply the distance dx by the slope f'(x) as in f'(x) Δx?

Yes, $dy$ is the vertical distance between two points on the tangent line and is equal by multiplying the horizontal distance $dx$ with $f'(x)$.

What do you mean by dy can be a number? In the figure from the book dy is the distance between the tangent and the graph and that distance is infinitesimal .

But it doesn't have to be an infinitesimal, it can be a bigger number too. All that $dy$ is defined as is the vertical distance on the tangent line. That can be infinitesimal or just finite and not infinitesimal. For example if $f(x) = x^2$ and if $x=1$, then we can choose $dx = 2$. Then $dy = f'(0)dx = 4$. So all that $dx$ and $dy$ are, are (hyperreal) numbers, where $dy$ depends on the number $dx$ of course.

Lastly just to make sure: Can you consider dy/dx as a number? because dy/dx = f'(x) and the slope f'(x) is surely a number.

Yes, $dy/dx$ is the quotient of two (hyperreal) numbers and thus a (hyperreal) number (it turns out to be a real number for the usual functions $f:\mathbb{R}\rightarrow \mathbb{R}$).

 

[christian0710]

. For example if f(x)=x2f(x) = x^2 and if x=1x=1, then we can choose dx=2dx = 2. Then dy=f′(0)dx=4dy = f'(0)dx = 4. So all that dxdx and dydy are, are (hyperreal) numbers, where dydy depends on the number dxdx of course.

I see, I think I'm starting to understand something i never understod: So for f(x)=x^2 and x=1, are you then saying that dx - the horizontal distance along the x-axis) is 2 full units long? and this gives f'(1)*dx =2*2 = 4 a corresponding vertical distance of 4 full units between the 2 points on the tangent?

Does that mean that the 2 points on the tangent are (1,2) and (3,4)
(1,2) because at x =1, y=x^2 = 2
(3,4) because you move 2 units out the x-axis, dx =2, and the slope of f'(1)=2 multiplied with the distance dx=2 is 4 :)

Interesting, i never thought that was possible.

 

[micromass]

I see, I think I'm starting to understand something i never understod: So for f(x)=x^2 and x=1, are you then saying that dx - the horizontal distance along the x-axis) is 2 full units long? and this gives f'(1)*dx =2*2 = 4 a corresponding vertical distance of 4 full units between the 2 points on the tangent?

Does that mean that the 2 points on the tangent are (1,2) and (3,4)
(1,2) because at x =1, y=x^2 = 2
(3,4) because you move 2 units out the x-axis, dx =2, and the slope of f'(1)=2 multiplied with the distance dx=2 is 4 :)

Interesting, i never thought that was possible.

Yes, but $dx$ can take other values as well. For example, with the same function, we can take $dx= 5$, then $dy = f'(1) dx = 10$.
The same can be done with $\Delta x$ and $\Delta y$. If $\Delta x = 5$, then $\Delta y = f(1 + \Delta x) - f(1) = f(6) - f(1) = 35$. Of course, we only really care about when $\Delta x$ is infinitesimal.

 

[christian0710]

Yess I understand it! so if we choose any point fx (1,f(1)), and set dx or dy to any number fx if dx=5, then dy=10 (dy depends on dx) in our example is interpreted as: When the slope of the tangent f'(1)=2 at point (1,f(1)) is multiplied by the horizontal run of 5 units, then we "run along the tangent" and end up at the point (1+dx, f(1) + dy) = (1+5, 2+10) = (6,12) ON the tangent.

So would this be the correct Conclusion: So dy and dx can be calculated for bigger numbers/values of dy and dx, and then we end up on a more distant point on the tangent. However, in practice this is not useful, because we are interested in using dy to approximate the slope of the function f at a point near f'(x). But why bother using this method if we have the derivative of a function? Is it not more practical to get the 100% accurate slope af a point as dy/dx = f'(x)?

 

[pondhockey]

Just a general note: doing "algebra" with dy/dx is only sometimes helpful, and other times is unsupported or unclear "magic" that substitutes a vague intuition for what is really intended. dy = f'(x) dx is an actual definition of dy. In this statement, dx and dy are both real numbers (not "infinitesimals"). The statement does not justify putting an integral sign in front of both sides of the definition.

I am now trying to learn thermodynamics; the equations of which are very often stated in "intuitive", non specific (mis) uses of differential notation. The equations often (or usually) leave out/hide functional dependencies that would imply use of the chain rule, and imply application of a path integral wrt a variable that is not indicated for example. At some point they become useful (IMO) only as mnemonics for those who already know the subject matter.

 

[christian0710]

Hi pondhockey, thank you for the reply. Could you give an example of when treating dx and dy as algebra seems like unsupported or unclear "magic" that substitutes a vague intuition for what is really intended. And would you be able to give an example of what the real way of doing it is? I think this could help me for when I'm going to read physics and chemistry.

 

[pondhockey]

Christian, here is an answer to one of my questions in a thermodynamics thread; I'll provide the "differential algebra magic" example later.

https://www.physicsforums.com/threa...ntials-in-thermodynamics.791811/#post-5135710

I'm "LATEX challenged" so I'll have to talk around it, but the link to Couchyam's answer is pretty explicit. Briefly, he expands an integral of PdV to what it really means. Note that it is NOT the integral of P with respect to V.

 

[pondhockey]

OK, here is an example of what I regard to be differential magic. I'm not going to attempt to provide the mathematically correct restatement because I'm still a thermodynamics grasshopper and I'll probably get it wrong. I think it conveys a "vague intuitive" understanding, but not a clear, precise understanding.

From H.Y. McSween, et. al. (2003) Geochemistry, pathways and processes. Columbia Press

Let P be pressure
let dV be a volume change.
let dq be an amount of heat gained.
T is (absolute) temperature
define dS = dq/T
let dE be a change in internal energy
write
dE + PdV<=TdS (p44; take this as given, so we can follow the "logic")

consider only processes for which dP = 0, and use the “fact” that

d(PV) = PdV + VdP (oh, really?! is d now the symbol for a derivative?)

to get

dE +d(PV) <= TdS

d(E+PV) <= TdS (so is this a new differential or a derivative?!)

etc.

 

[christian0710]

Thank you for sharing, when I', done with calculus and start reading Thermodynamics, I'll return to this post and it will probably make more sense to me :)

 

[Mark44]

OK, here is an example of what I regard to be differential magic. I'm not going to attempt to provide the mathematically correct restatement because I'm still a thermodynamics grasshopper and I'll probably get it wrong. I think it conveys a "vague intuitive" understanding, but not a clear, precise understanding.

From H.Y. McSween, et. al. (2003) Geochemistry, pathways and processes. Columbia Press

Let P be pressure
let dV be a volume change.
let dq be an amount of heat gained.
T is (absolute) temperature
define dS = dq/T
let dE be a change in internal energy
write
dE + PdV<=TdS (p44; take this as given, so we can follow the "logic")

consider only processes for which dP = 0, and use the “fact” that

d(PV) = PdV + VdP (oh, really?! is d now the symbol for a derivative?)

I don't see anything wrong here. "d" means "differential of", not "derivative of".

to get

dE +d(PV) <= TdS

d(E+PV) <= TdS (so is this a new differential or a derivative?!)

Again, d here means differential. The operator is linear, so the differential of a sum of functions is the sum of the differentials of the functions.