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Problem understanding mathematical argumentation please help

  1. Jun 23, 2015 #1
    Hi, my Calculus book (Elementary Calculus an Infinitesimal approach) uses a mathematical argument, and I'm not sure if it's correct so i thought i wuold check with you guys :)

    When the book want's to find the derivative of f(x)=x^2 the books starts like this

    y=x2
    Δy +y = (x + Δx)2

    So my question is this: Is it mathematically correct to write
    Δy+y = (x + Δx)2.

    To me it looks more like the left side should translate into a right side that looks like this
    Δy+y = x2 + (Δx)2

    I'm asuming that Δy+y translates into f(Δx+x) and not f(x)+f(Δx) but to me it looks as if Δy+y is f(x)+f(Δx) which is not the same as f(Δx+x)

    If the above is true, does that mean that this is also true?

    if y=x2 then.
    y=x2
    y+1/y = (x+1/x)2 and would this be the same as f(x+1/x)?





     
    Last edited by a moderator: Jun 23, 2015
  2. jcsd
  3. Jun 23, 2015 #2
    No, here x2 is a function of x
    That is y=f(x)=x2
     
  4. Jun 23, 2015 #3

    Mark44

    Staff: Mentor

    Yes.
    No, not at all. ##(x + \Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2##, not x2 + (Δx)2 as you have.
    Δy+y = f(Δx+x) ≠ f(Δx) + f(x).
    You are mistakenly thinking that the distributive property applies here, which it does not.
    The distributive property applies when you multiply a sum of numbers, as in a(b + c) = ab + ac. It does NOT apply to functions, in general. For example, ##\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}##, and ##\sin(x + y) \neq \sin(x) + \sin(y)##.
    Let's define y = f(x) = x2

    What is y + 1/y?
    What is f(x + 1/x)?
    Are they equal?
     
  5. Jun 23, 2015 #4
    Del means something like a little change of..

    So here a little change of x is added. For example, when you put x=2 then y = 4,
    With a little change in x like del x=1 then x will be equal to 5, and y will be equal to 52=25, not (4)2+(1)2
     
    Last edited: Jun 23, 2015
  6. Jun 23, 2015 #5
    Thank you for the reply, really appreciate it. So to answer your question:
    if y = f(x) = x2
    Then y+1/y = (x+1/x)^2 which is the same as f(x+1/x) Correct?

    So if i take f(f(x+1/x)) is it then the same as f(x+1/x) = (x+1/x)^2 ? or is it ( (x+1/x)+1/(x+1/x))2?
     
  7. Jun 23, 2015 #6

    Mark44

    Staff: Mentor

    No. Let's start from the other end.
    f(x + 1/x) = (x + 1/x)2, which is what you have.
    But (x + 1/x)2 = x2 + 2 * x * 1/x + 1/x2 = x2 + 2 + 1/x2 = f(x) + 2 + f(1/x)

    The last expression is not equal to either y + 1/y or f(x) + f(1/x).
     
  8. Jun 24, 2015 #7
    Then I think I'm misunderstanding one very important thing, so let's see where my logic is wrong:

    Above we agreed that
    Δy+y = f(Δx+x), So i assume this means "So if you add some y or Δy or 1/y to the left side it's the same as putting it into the f(x) operator, so if the left side says
    Δy+Δy+y then this is equal to f(Δx+Δx+y) and therefor it should also be true that y+1/y = f(x+1/x) So I'm erroneously treating Δy like y?


    Just to make sure, could this be one of my wrong ways of thinking ?
    If f(x)=y=1/x then
    y+Δy = 1((x+Δx) which is true BUT now I'm erroneously thinking that if you put something on the left side, it goes into the "x" on the right side, and therefore:
    y+y = 1/(x+x) (I think I'm treating γ like Δy) while in fact it should be y+y = 1/x+1/x =2/x

    So my question is, why don't we treat Δy and y the same way? Why is it not Δy+y = 1/x +1/Δx

    UPDATE: Is this because Δy means "An increment in y" while y means "A complete function", so by adding 2*Δy on the left side we are not adding 2 functions on the right, but we are incrementing x to 2*Δx and this is the same as (x + 2*Δx) :D
     
    Last edited: Jun 24, 2015
  9. Jun 24, 2015 #8

    Mark44

    Staff: Mentor

    In general, no.

    In your first post, you started with f(x) = x2, so lets stick with that.
    You should sketch a graph of this function, with two points labeled: A(x, x2) and B(x + Δx, (x + Δx)2). The horizontal distance between A and B is Δx. The vertical distance between A and B is Δy = (x + Δx)2 - x2 = 2xΔx + (Δx)2. If we add a third point C(x + 2Δx, (x + 2Δx)2), the vertical distance between B and C will NOT be Δy.

    BTW, f is a function, not an operator. f(x) is the number that the function associates with a value x in the domain of f.
    Please reread what I wrote in post # 6. I already addressed why this is wrong.
    No, it's not true, due to a typo.

    Since you started off this thread with f defined as f(x) = x2, you should use a different letter if you now want to work with a different function.

    Let's say g(x) = 1/x
    So g(x + Δx) = ##\frac{1}{x + \Delta x}##
    ##\Delta y = g(x + \Delta x) - g(x) = \frac{1}{x + \Delta x} - \frac{1}{x}##
    Δx is a symbol that represents a "small" change in x. Δy similarly represents a small change in y. More precisely, Δy = f(x + Δx) - f(x).

    The main problem, as I see it, is that you are focussing on the left side without understanding what the right side of your equation means.
     
  10. Jun 24, 2015 #9

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're getting yourself in a real old tangle here! First (by definition):

    ##\Delta y = y(x + \Delta x) - y(x)##

    And that's about it, really.

    Whereas, I think you want to define:

    ##\Delta y = y(\Delta x)##
     
    Last edited: Jun 24, 2015
  11. Jun 24, 2015 #10
    It's something like y=f(x), so, /frac{1}{y}=/frac {1}{f(x)}

    In a function you should put the value of the variable of which the function is of. In case of f(x) for new values of x (which may be x+del x, or 1 or 3 or any such) you should put the value in the function.
     
  12. Jun 24, 2015 #11
    +
    YES I think I finally understand it now, I just couldn't see the full argument : Do we agree on this:

    The book says f(x) = x2

    So when we start with
    y = y(x) =x^2

    and when we add Δy, we are actually doing this

    y +Δy = y(x) +(y(x+Δx) -y(x)) = y(x+Δx) = (x+Δx)2

    I couldn't understand that when we add Δy on the left side, it looked like it suddently appeared inside y(x) as y(x+Δx). beucase all i saw on the left side was, that when you add Δy on the left side, it appears inside x as (x+Δx) so i thought that because Δy+y = f(x+Δx) then it must also be true that y+y = f(x+x) but Mark44 helped me se that it's not the case and the reason that y +Δy = y(x+Δx) is because y +Δy = y(x) +(y(x+Δx) -y(x)) = y(x+Δx)

    Thank you guys :D
     
  13. Jun 26, 2015 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The point is that they are trying to answer "if we increase x a small amount, what happens to y?" If [itex]y= x^2[/itex] and we increase x by the small amount [itex]\Delta x[/itex], then y is changed to [itex](x+ \Delta x)^2= x^2+ 2(\Delta x)x+ (\Delta x)^2[/itex]. Since that [itex]x^2[/itex] is just y itself, this is the same as [itex]y+ 2(\Delta x)x+ (\Delta x)^2[/itex]. So increasing x by [itex]\Delta x[/itex] increases y by [itex]2(\Delta x)x+ (\Delta x)^2[/itex] which, since it is a small increase in y, we call [itex]\Delta y[/itex].
     
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