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Proving/deriving first fundamental theorem of calc

  1. May 6, 2013 #1
    I know this is going to be atrociously bad, but I like to try to prove things.

    [itex]\frac{f(x+\Delta x) - f(x)}{ \Delta x} = \frac{\Delta y}{\Delta x}[/itex]

    [itex] => f(x+\Delta x) - f(x) = \frac{\Delta y}{\Delta x} \Delta x[/itex]

    [itex] => f(x+\Delta x) = \frac{\Delta y}{\Delta x} \Delta x + f(x)[/itex]

    Now, express x as (x2 + delta x) and define f(x) in the same way we just defined f(x+deltax)

    [itex] => f(x+\Delta x) = \frac{\Delta y}{\Delta x} \Delta x + \frac{\Delta y}{\Delta x} \Delta x + f(x2)[/itex]

    We could now express x2 as (x3 + delta x) and define f(x2) in a similar way again.

    [itex] => f(x+\Delta x) = \frac{\Delta y}{\Delta x} \Delta x + \frac{\Delta y}{\Delta x} \Delta x + \frac{\Delta y}{\Delta x} \Delta x + f(x3)[/itex]

    (I should have subscripts or something similar, because these delta y/delta x's are not all the same thing, but bear with me :)

    It is clear that repeating this process over and over becomes a sum of all delta x/ delta y

    [itex] => f(x+\Delta x) = \sum \frac{\Delta y}{\Delta x} \Delta x[/itex]

    If we take the limit of both sides as delta x approaches 0, the left side becomes f(x) and the right side becomes the definition of the integral.

    [itex] f(x) = \int f'(x) dx[/itex]

    I know a lot of things are messed up about it, I think I need to name all my different x's and y's for starters, and I'm sure I said a lot of things that didn't make sense as well, but this is a 10 minute attempt.

    Any input is appreciated.
    Last edited by a moderator: May 8, 2013
  2. jcsd
  3. May 6, 2013 #2


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    You have the general idea.
  4. May 6, 2013 #3
    That's quite surprising.
  5. May 6, 2013 #4

    Stephen Tashi

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    Doesn't that demonstration use [itex] \triangle y [/itex] to stand for more than one quantity? Is [itex] \triangle y [/itex] going to be [itex] f(x_2) - f(x_1) [/itex] or is it going to be [itex] f(x_3) - f(x_2) [/itex] etc. ? I think you need subscripts to distinguish the [itex] \triangle y [/itex]'s.

    I think your idea amounts to proving the fundamental theorem of calculus from an agrument using the Calculus of Finite Differences.
  6. May 7, 2013 #5
    Yes, as noted those delta y's are all different. I definitely need subscripts, I'll try to rewrite it sometime today.

    I don't know what the Calculus of Finite differences is, but I'll check it out. My direction was just based off "the derivative is the quotient of a difference, and the integral is the sum of a product." Knowing that that was inverse-like, I figured that solving for f(x+h) in the derivative definition could give me something integral-like.

    The only hurdle is that the derivative is "of a point" while the integral isn't, which I'm guessing is the reason why the sum of all of the delta y/delta x makes an appearance, as opposed to two terms as in the derivative.
    Last edited: May 7, 2013
  7. May 7, 2013 #6

    Stephen Tashi

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    A basic idea in the Calculus Of Finite Differences is the relation between summation and differencing.
    Let [itex] h [/itex] be a given constant. For a function [itex] G(x) [/itex] , define the function [itex] F [/itex] whose domain will be the integers by [itex] F(k) = G(kn).[/itex].

    Define the function [itex] f(k) = \triangle F(k) = F(k+1) - F(k) [/itex]

    If we face the problem of doing a summation [itex] S = \sum_{i=1}^n f(n) [/itex] then this amounts to the easy "telescoping" sum:

    [itex] S = f(1) + f(2) + .. .+ f(n) = (F(2)-F(1)) + (F(3)-F(2)) + ... +(F(N+1)-F(N)) [/itex]
    [itex] = F(N+1) - F(1) [/itex]

    One can approach doing finite a summation of a function [itex] f [/itex] as a process of seeking an "anti-difference" function for [itex] f [/itex].
  8. May 7, 2013 #7

    Stephen Tashi

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    I meant [itex] F(k) = G(kh) [/itex] of course.
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