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I Why dy/dx is not a ratio?

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  1. Mar 6, 2017 #1
    I read people saying that dy/dx is not a ratio because it is a limit or standard part of a ratio. $${dy\over dx} = \lim_{h \to 0} {f(x +h) - f(x) \over h}, \ \ \ {dy\over dx} = st\left( {f(x +h) - f(x) \over h}\right)$$

    what I get is ##{f(x +h) - f(x) \over h}## is a ratio but putting a limit or st makes it something else than a ratio.

    So my question is ##\lim_{x \to 0} \frac 4 5## a ratio ? If it is then how would you put an formal argument against dy/dx as a ratio ?
     
    Last edited: Mar 6, 2017
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  3. Mar 6, 2017 #2

    PeroK

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    I was going to answer, but first define "ratio".
     
  4. Mar 6, 2017 #3
    I never thought about this and honestly I don't know the definition. Let me try anyways,

    We say two numbers ##p, q## are in ratio if we can find a third number ##r## such that ##p = rq##.
     
  5. Mar 6, 2017 #4

    PeroK

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    Unless ##q = 0##, numbers are always in a ratio. I was expecting you to say something like a ratio is one number divided by another. In particular:

    ##\frac{f(x_0+h) -f(x_0)}{h}## is a ratio of two numbers.

    But, as ##dx## and ##dy## are not numbers, ##\frac{dy}{dx}## is not a ratio of two numbers.

    That's important because you may have proved that for any (non-zero) numbers, you have: ##\frac{a}{b} \frac{b}{c} = \frac{a}{c}##.

    But, it doesn't immediately follow that ##\frac{dy}{dx} \frac{dx}{dz} = \frac{dy}{dz}##.

    Instead, you have to prove this (as separate and additional to your proof for numbers) using the definition of these things as limits.

    From a pure maths point of view, making things look like already familiar things doesn't imbue them with the same properties; those properties must be proved in themselves.
     
  6. Mar 6, 2017 #5
    Why ? Lets say ##{dy \over dx} = {f(x+ h) - f(x)\over h}## then it is a ratio as ##{f(x+ h) - f(x)\over h}## is a ratio. But ##{dy \over dx} = \color{red}{\lim_{h \to 0}} {f(x+ h) - f(x)\over h}## then is not a ratio. So it must be something with the limit right ?

    So if putting a limit make things something other than ratio then ##\lim {4\over 10}## is not a ratio. How will we justify ##\lim {4\over 10}## as a ratio ?
     
  7. Mar 6, 2017 #6

    PeroK

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    This is such a confusion of ideas, it's hard to answer. ##\frac{4}{10}## is no more a ratio than any other number. It's ##0.4##. It's just a plain old number. Like all numbers, it can be written as a ratio in an infinite number of ways.

    You can't make something a number by wishing it to be a number. How would you define, for example, ##dx## as a number? What number is it for a given function? Let's say the simple function ##f(x) = x^2##. What is ##dx## at ##x=1##?
     
  8. Mar 6, 2017 #7

    PeroK

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    Since this isn't homework, let me explain the problem. ##dx## is an "infinitesimal" change in ##x## and likewise ##dy##. Now, infinitesimals are not numbers. Why? Well, a number is either ##0## or it isn't. And, if you take an infinitesimal as so small that it's 0, then all you have is ##dx = dy = 0## in all cases (for all functions at all points). And that's no good.

    But, if ##dx## is not ##0##, then it's just some finite number. But, then, it's just some finite change in ##x##. To keep things simple, let's assume that ##dx = 1##. If we take the function ##f(x) = x^2## we have:

    ##f(x+1) - f(x) = (x+1)^2 - x^2 = 2x + 1##

    So, for this function, we have ##dy = 2x + 1## and ##\frac{dy}{dx} = 2x + 1##.

    And that's no good either. That's just an approximation of the derivative. If we took ##dx## smaller, we would get a better approximation. But, we can never get ##\frac{dy}{dx} = 2x##.

    In fact, if ##dx## is a number we get ##\frac{dy}{dx} = 2x + (dx)^2##.

    And that's the problem: an infinitesimal like ##dx## can't be ##0## and it can't be non-zero, thus it can't be a number.

    In fact, in a formal development of derivatives, you would tend to use ##f'(x)## for the derivative. And, once you proved the properties of the derivative and the integral, especially in applied maths, you tend to adopt the ##\frac{dy}{dx}## notation, because it can be easier to work with in things like the chain rule and the fact that you can cancel infinitesimals as though they were numbers is of considerable appeal.

    But, never at any stage do ##dx## and ##dy## become real numbers.
     
  9. Mar 6, 2017 #8
    @PeroK Ok I get your point that dy and dx are not numbers and so dy/dx is not a ratio but lets forget that we denote dy/dx for a derivative. Lets say we write the first principle for derivative everytime we differentiate, like instead of ##{d(x^2)\over dx} = 2x## we write ##\lim_{x \to 0} {(x+h)^2 - x^2\over h} = 2x##. Now there is no dy or dx to worry about.

    We also did not change anything other than we replaced the shorthand for derivate with its definition. Now it is still not a ratio of any sort. The ##\lim## part is what disabling it to be a ratio.

    We apply the same argument to the things like ##\lim 4/3## or whatever number you like. By that logic we say this is also not a ratio, but then there is nothing like a ratio because we can just replace 4/3 with anything and argument still holds the same. This conclusion can be easily contradicted with the fact that ##4 = 2 \times 2## thus ##4:2 = 2:1 ## is a ratio.

    So what went wrong in my thinking ?

    Sorry I think I did not have phrased the question properly, the real problem that is bugging me is that we say ##\lim_{\delta x \to 0} {f(x + \delta x) - f(x) \over \delta x}## is not ratio but we say ##\lim 4/3## as a ratio.
     
  10. Mar 6, 2017 #9

    PeroK

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    You need to put aside the idea that there are special numbers called "ratios". The limit is the limit of a numerical function (it's actually a function of ##h## with ##x## constant while you take the limit). So:

    ##f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}##

    There is nothing special about numbers that are ratios. This is just a functional limit and should be understood as such.
     
  11. Mar 6, 2017 #10
    Sorry I can relate this to my previous post. Can you dumb this down a bit ? please.
     
  12. Mar 6, 2017 #11

    PeroK

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    Perhaps you are not ready for real analysis. It's not the easiest thing to pick up and grasp. Are you a maths student?
     
  13. Mar 6, 2017 #12
    Yes.
     
  14. Mar 6, 2017 #13

    PeroK

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    Have you done limits yet?
     
  15. Mar 6, 2017 #14
    @PeroK
    Yes (I once answered a question about limits) . I know multivariable calculus that is required to do electrodynamics.
    I have also asked some questions about series, so I have done some real analysis at least.
    But I have never learnt that a dy/dx is not a ratio, so I was a bit confused. Sorry if I troubled you much.
     
    Last edited: Mar 6, 2017
  16. Mar 6, 2017 #15
    You will need to learn about differential forms to understand what dx and dy is. I suggest either Spivak's Calculus on Manifolds or Munkres' Analysis on Manifolds.
     
  17. Mar 6, 2017 #16

    Math_QED

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    ##\frac{dy}{dx}## should not be interpretated as a ratio, but simply as notation for the derivative. The notation is useful because in some cases it corresponds with our intuition when we manipulate fractions. For example, without mentioning the conditions when this is true:

    ##\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}##.

    If ##f: (a,b) \rightarrow \mathbb{R}##, then ##\frac{df}{dx}: (a,b) \rightarrow \mathbb{R}: x \mapsto \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}##

    if these limits exist for all ##x \in (a,b)## (otherwise we must change the domain)
     
    Last edited: Mar 6, 2017
  18. Mar 6, 2017 #17
    Yes I know, I have the same definition in my high school book.
     
  19. Mar 6, 2017 #18

    Math_QED

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    Then what is your exact problem?
     
  20. Mar 6, 2017 #19
    Did you read post #8 ?

    But I think I lack knowledge to understand this now.
     
  21. Mar 6, 2017 #20

    Drakkith

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    PeroK, if I understand correctly, you're saying that limits only apply to functions?
     
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