Some questions about notation

Discussion in 'Math' started by Lightfire, Apr 3, 2014.

  1. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    I know this will be another simple question of mine, but I just can't help it.

    Often in integration, we use substitution like

    u=2x

    Then du=2 dx

    Okay, I understand it in intuitive sense. I also understand why 2x became 2.

    Now my question is

    1. Why du is no fraction like \frac{du}{dx}?

    2. Why 2 dx has dx on it? When we differentiate things like: We have a function y=3x^2, then the notation and result would be \frac{dy}{dx}3x^2=6x, right? There isn't dx on the right side of the equation.

    My guess to my question is this:

    Actually, suppose we have u=2x, then we differentiate it \frac{du}{dx}2x=2 Then we multiply dx (I guess this is very informal because I treat as if dx is a number when in fact \frac{du}{dx} is just a notation.) to both sides of the equation.

    \frac{du}{dx}dx=2dx

    du=2dx

    I know I have removed the 2x in \frac{du}{dx}2x=2. But as of now, that is how I understand it.

    I know my guess is very informal so I hope the geniuses here will help me.

    Thank you.
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    I don't have the full answer for you and I'm away from my reference, but treating differentials like dy and dx separately works OK if the functions are well behaved. In this context it means continuous and integrable. When functions get pathological many informal methods breakdown.
     
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  3. atferrari

    AAC Fanatic!

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    That's serious PB! :eek: :p
     
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  4. amilton542

    Active Member

    Nov 13, 2010
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    1) Because you've separated the variables into differentials.

    2) No.

     \frac{d}{dx} 3x^2 = 6x

     (d)(3x^2) = (6x)(dx)
     
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  5. amilton542

    Active Member

    Nov 13, 2010
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    If you let u = 3x^2, then you'll see:

     du = f'(x)dx

    The change in u accelerates the change in x.
     
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  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    First you select the quantity that you want to be u.

    In your case you want 2x to be u. So, u=2x.

    Then you take the DERIVATIVE of u with respect to x. Get it? du=2dx is not something that you pull out of air or magic. You have to take DERIVATIVE of u with respect to x. So, derivative of u with respect to x is derivative of 2x with respect to x:
    \frac{d}{dx}u=\frac{d}{dx}2x=2\frac{d}{dx}x=2(1)=2

    \frac{d}{dx}u=2

    \frac{du}{dx}=2

    du=2dx

    The only thing YOU choose is which "thing" will be u. Afterward everything is mathematically derived.
     
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  7. WBahn

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    Mar 31, 2012
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    Keep in mind that the notation dy/dx isn't arbitrary.

    It is the ratio of a small change in y, Δy, to a small change in x, Δx, as the changes get arbitrarily small. So fundamentally you are working with something like:

    y = 2x²

    Δy/Δx ≈ 4x

    It then makes perfect sense to multiply both sides by Δx to get

    Δy ≈ 4x*Δx

    What this says is just that if we change x by a small amount, the change in y is 4 time that amount times the value of x. If x is twice as big, then the change in y will be twice as big.

    The approximation becomes equality in the limit that we are talking about infinitesimal changes.

    dy = 4xdx

    but the interpretation is essentially the same.
     
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  8. MrChips

    Moderator

    Oct 2, 2009
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    You are being too abstract and forgetting what derivatives and differentiation mean.

    Differentiation means taking the slope or gradient, i.e. Δy/Δx

    Let us start with a simple example:

    y = 4x

    Graphically, this can be represented by a straight line.
    What is the slope of this line? We know the slope of a straight line is a constant value.

    Hence we can use Δy/Δx.

    y = 0 when x = 0
    y = 4 when x = 1
    Δy/Δx = (4 - 0)/(1 - 0) = 4/1 = 4

    Δy/Δx = 4 for all values of x

    Now let us take

    y = 4x²

    We know this is not a straight line.
    What is the slope of the line? It is not constant. It varies at every position of x.

    Can use Δy/Δx to find the slope? Yes, we can find the slope at any point x as long as we make Δx very small at the point x.

    For y = 4x²

    dy/dx = 8x

    This is the same as saying

    Δy/Δx = 8x in the limit as Δx approaches 0.

    Thus dy/dx is simply a way of expressing the slope or gradient of a function at any value of x.


    dy = 8x . dx is the same equation as dy/dx = 8x

    just as writing

    Δy/Δx = 8x is the same as Δy = 8x . Δx

    Note that neither dx nor Δx is zero.
     
    Last edited: Apr 5, 2014
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  9. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Thank you everyone!

    I can't help myself but can you give an example? Please?:D



    y = 2x²

    Δy/Δx ≈ 4x

    Let's say that we choose x to be 0.001 so the y will be 0.000002.

    So we have Δy/Δx as 0.000002/0.001 = 0.002 but that's not equal to 0.004 (or 4*0.001)

    Isn't it equal because im wrong or because it can't be equal because that's just an approximation?

    Or my interpration is wrong?

    P.S.: If it would be okay pls help me with http://forum.allaboutcircuits.com/showpost.php?p=699874&postcount=19 ... I'd appreciate everyone!

    Thanks!
     
  10. MrChips

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    Oct 2, 2009
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    y = 2x²

    Δy/Δx ≈ 4x

    At x = 0.001, y = 0.000002

    Δy is not 0.000002
    Δx is not 0.001

    At x = 0.001, y = 0.000002
    At x = 0.0011, y = 0.00000242

    Δx = 0.0001, Δy = 0.00000042
    Δy/Δx = 0.00000042/0.0001 = 0.0042

    dy/dx at x = 0.001 = 4 times 0.001 = 0.004
     
    Last edited: Apr 5, 2014
  11. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    touché!

    That's why earlier, I think there was an error in my post. I thought Δy/Δx must be like change in over change in x....

    I really learn many things here. The things I think I already know, experts here falsify that! That's why I love asking here, eh. People's not bully.:)

    Thanks...

    Can someone help me with my Fourier problem? I'd like to understand it in mathematical sense, at least... thanks!
     
  12. MrChips

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    What's the Fourier problem?
     
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  13. WBahn

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    Δy/Δx IS change in y over change in x.

    Let's again consider

    y = 2x²

    Pick any x, say x = 10

    y = 200

    Now pick a small change in x, say Δx = 0.01

    y at x=10.01 = 200.4002

    So Δy = 0.4002

    Δy/Δx = 0.4002/0.01 = 40.02

    dy/dx = 4x = 4*10 = 40
     
  14. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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  15. WBahn

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  16. MrChips

    Moderator

    Oct 2, 2009
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    Don't paste link into your reply box. Use the Insert Link button (icon) at the top of the message box.
     
  17. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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