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F(x+dx)=F(x)+F'(x)dx eqn an identity

  1. Jun 8, 2005 #1

    is the above eqn an identity or something?can someone explain to me what is happening in this eqn?
    Last edited: Jun 8, 2005
  2. jcsd
  3. Jun 8, 2005 #2


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    It's the approximation which follows from the definition of the derivative

    [tex] F'(x)=:\lim_{\Delta x\rightarrow 0} \frac{F(x+\Delta x)-F(x)}{\Delta x} [/tex]

    by not considering the limit,but the increment in "x",viz.[itex] \Delta x [/itex],is very,very small (infinitesimal,if you prefer).

  4. Jun 9, 2005 #3
    Locate a discussion of Taylor series and the Newton-Rapson algorithm. You'll find a good bit on both in a Numerical Analysis textbook. Now, if you are reading a NA text, and that makes you ask your question, then I am sorry for redirecting you to where you already are. In that case, just notice the obvious.

    If dx = 0, then of course f(x+dx) = f(x) + f'(x)dx.

    What happens if we move dx slightly away from zero? ...well, see dextercioby's last post.
    Last edited: Jun 9, 2005
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