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Thanks in advance for any help.

PS: I'm sorry for my bad english. .

- Thread starter DeathKnight
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- #1

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Thanks in advance for any help.

PS: I'm sorry for my bad english. .

- #2

matt grime

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It isn't a fraction because, well, it isn't a fraction. Crudely, a fraction is a ratio of two numbers*. a/b, and neither dy nor dx are numbers. But, you can under certain special circumstances treat it like a fraction. It is lazy, but perfectly acceptable to do so for practical purposes and all you're doing is this:

rearrange your DE if possible so that

f(y)dy/dx=g(x)

for some functions of f and g.

Now, integrate both sides with respect to x

[tex]\int f(y)\frac{dy}{dx}dx = \int g(x)dx[/tex]

but the one on the left is exactly the same as

[tex] \int f(y)dy[/tex]

so you can *pretend* that the dx's cancel out. Technically they do cancel out but *not* because it's a fraction, but because of what the symbols mean as calculus tools.

* properly the symobl 1/x ought to be interpreted as the multiplicative inverse of x (which is why 1/0 is not defined), and as the dx's are not invertible in this sense that is why we don't call them fractions.

rearrange your DE if possible so that

f(y)dy/dx=g(x)

for some functions of f and g.

Now, integrate both sides with respect to x

[tex]\int f(y)\frac{dy}{dx}dx = \int g(x)dx[/tex]

but the one on the left is exactly the same as

[tex] \int f(y)dy[/tex]

so you can *pretend* that the dx's cancel out. Technically they do cancel out but *not* because it's a fraction, but because of what the symbols mean as calculus tools.

* properly the symobl 1/x ought to be interpreted as the multiplicative inverse of x (which is why 1/0 is not defined), and as the dx's are not invertible in this sense that is why we don't call them fractions.

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- #3

mathwonk

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they are frequently defined as just one indivisible quantity because it is some work to define them separately.

they can however be defiend as separate entities, and this is sueful in diffrential equations.

for this reason some good diff eq books do the work required to define them carefully and separately, such as tenebaum and pollard, but most just skip it and use them separately anyway.

As Matt points out, using them separately does not actually require defining them separately.

the way of defining their quotient treats each as a funcrtion into a one dimensional vector space. then the quotient is function whose value is a quotient of vectors in the same one dimensional space. such vector can be divided if and only if the bottom one is not zero.

i.e. given any non zero vector v in a one dimensional vector space, there is a unique number c associated to each other vector w such that w = cv. in that sense c = w/v.

if one is dealing with analytic or algebraic objects, even if the function on the bottom is zero a few places, those places are isolated, and the quoptient still makes sense as a meromorphic or rationalk function.

but i think you don't really need to know all this to use the technique, and this is why it is usually not explaained in detail. you might look at tenenbaum and pollard though for a simpler explanation if you are still curious..

they can however be defiend as separate entities, and this is sueful in diffrential equations.

for this reason some good diff eq books do the work required to define them carefully and separately, such as tenebaum and pollard, but most just skip it and use them separately anyway.

As Matt points out, using them separately does not actually require defining them separately.

the way of defining their quotient treats each as a funcrtion into a one dimensional vector space. then the quotient is function whose value is a quotient of vectors in the same one dimensional space. such vector can be divided if and only if the bottom one is not zero.

i.e. given any non zero vector v in a one dimensional vector space, there is a unique number c associated to each other vector w such that w = cv. in that sense c = w/v.

if one is dealing with analytic or algebraic objects, even if the function on the bottom is zero a few places, those places are isolated, and the quoptient still makes sense as a meromorphic or rationalk function.

but i think you don't really need to know all this to use the technique, and this is why it is usually not explaained in detail. you might look at tenenbaum and pollard though for a simpler explanation if you are still curious..

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I had a prof in Analysis that was very concerned with the non-existance of infinitesimals, and did not want to see dx used even for integration. However, about that time Conway came up with surreal numbers. Nevertheless, from what i have been reading they are not useful in integration:

* it's proving hard to integrate with surreals - crudely, to add up *

infinite amounts of infinitely small quantities and get sensible

results. Without integration, surreals lose much of their interest

in physics..

However, article goes on to say:* Another way to represent infinitesimals as an extension of the real numbers is by Robinson's Nonstandard Analysis. Robinson's Nonstandard numbers are a subfield of the Surreals.* http://www.valdostamuseum.org/hamsmith/surreal.html [Broken]

infinite amounts of infinitely small quantities and get sensible

results. Without integration, surreals lose much of their interest

in physics..

However, article goes on to say:

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- #5

Hurkyl

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Let std(x) denote the "standard part" of a finite number -- that is, you round the number to the nearest standard number.

Or, to put it a different way, if the quantity

[tex]

\text{std} \left( \frac{\Delta y}{\Delta x} \right)

[/tex]

is the same no matter what nonzero infinitessimal difference you use, then

[tex]

\frac{dy}{dx} = \text{std} \left( \frac{\Delta y}{\Delta x} \right)

[/tex]

- #6

Integral

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My slant is that [itex] \frac {dy} {dx} [/itex] is notational short hand for:

[tex] \frac d {dx} y(x) [/tex]

Where

[tex] \frac d {dx} [/tex] is the notation framework indicating that the process is differentiation and that the independent variable is x, so the process to be done is differentiation wrt x. The object of this process will be the function y(x). Note that x is the independent variable of the function y, this must agree with the independent variable indicated in the [itex] \frac d {dx} [/itex] notation.

edit for latex check

So why is it not a fraction? Because it is notation. We are very fortunate that the notation is flexible enough to allow fraction like operations when needed. That however, does not make it a fraction.

[tex] \frac d {dx} y(x) [/tex]

Where

[tex] \frac d {dx} [/tex] is the notation framework indicating that the process is differentiation and that the independent variable is x, so the process to be done is differentiation wrt x. The object of this process will be the function y(x). Note that x is the independent variable of the function y, this must agree with the independent variable indicated in the [itex] \frac d {dx} [/itex] notation.

edit for latex check

So why is it not a fraction? Because it is notation. We are very fortunate that the notation is flexible enough to allow fraction like operations when needed. That however, does not make it a fraction.

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matt grime

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"What about the chain rule ? what happens there if the two infinitesimals do not cancel ?"

Then you are probably not using the chain rule (properly). The chain rule states that for a z a function of y and y a function of x, then

[tex] \frac{dz}{dy}\frac{dy}{dx}=\frac{dz}{dx}[/tex]

and it is easy to prove this result if one instead uses the 'dash' notation and the 'o' notation properly. Modulo rigorousness it states that if

f(y+h) ~ f(y)+ f'(y)h

and if g(x+h) ~ g(x)+hg'(x)

then f(g(x+h)) ~ f(g(x)+hg'(x)) ~ f(g(x) + hg'(x)f'(g(x))

Then you are probably not using the chain rule (properly). The chain rule states that for a z a function of y and y a function of x, then

[tex] \frac{dz}{dy}\frac{dy}{dx}=\frac{dz}{dx}[/tex]

and it is easy to prove this result if one instead uses the 'dash' notation and the 'o' notation properly. Modulo rigorousness it states that if

f(y+h) ~ f(y)+ f'(y)h

and if g(x+h) ~ g(x)+hg'(x)

then f(g(x+h)) ~ f(g(x)+hg'(x)) ~ f(g(x) + hg'(x)f'(g(x))

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Hello DeathKnight !PS: I'm sorry for my bad english.

Apparently, we are in the same boat... I don't know if you are French, but if so or if you can read French papers, this is a review for general (French) public about dx and related notations :

"Une querelle des Anciens et des Modernes"

http://www.scribd.com/JJacquelin/documents

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