Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    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+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


    Staff: Mentor

    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


    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


    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


    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


    User Avatar
    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].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook