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Math Help - Big-O Notation in Geometric Interpretation of Derivative

  1. #1
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    Big-O Notation in Geometric Interpretation of Derivative

    I just watched the first lecture of , and in that video, the lecturer explained about the geometric interpretation of derivative. A derivative of a function f(x) can be defined as:

    f'(x) = \lim_{x \to \Delta x} \frac{f(x+\Delta x) - f(x)}{\Delta x}

    The lecturer then used this to find the derivative of the function f(x) = x^n, and this is where it starts to get confusing for me. Using the definition above, the derivate of the function f(x) = x^n is equivalent to:

    f'(x) = \lim_{x \to \Delta x} \frac{(x+\Delta x)^n - x^n}{\Delta x}

    The part which confuses me is when the term (x+\Delta x)^n is evaluated using the Binomial Theorem:

    (x+\Delta x)^n = x^n + n(\Delta x)x^{n-1}+\mathit{O}((\Delta x)^2)

    Why are the rest of the terms replaced by the Big-O Notation \mathit{O}((\Delta x)^2)? And why the term (\Delta x)^2 specifically?

    Thanks!
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  2. #2
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    Quote Originally Posted by fishcake View Post
    I just watched the first lecture of , and in that video, the lecturer explained about the geometric interpretation of derivative. A derivative of a function f(x) can be defined as:

    f'(x) = \lim_{x \to \Delta x} \frac{f(x+\Delta x) - f(x)}{\Delta x}

    The lecturer then used this to find the derivative of the function f(x) = x^n, and this is where it starts to get confusing for me. Using the definition above, the derivate of the function f(x) = x^n is equivalent to:

    f'(x) = \lim_{x \to \Delta x} \frac{(x+\Delta x)^n - x^n}{\Delta x}

    The part which confuses me is when the term (x+\Delta x)^n is evaluated using the Binomial Theorem:

    (x+\Delta x)^n = x^n + n(\Delta x)x^{n-1}+\mathit{O}((\Delta x)^2)

    Why are the rest of the terms replaced by the Big-O Notation \mathit{O}((\Delta x)^2)? And why the term (\Delta x)^2 specifically?

    Thanks!
    What big-O means in this case is that there exists a positive constant K (which may depend on x )such that for |\Delta x| small enough:

    <br />
\left|(x+\Delta x)^n - [x^n + n(\Delta x)x^{n-1}]\right| < K |\Delta x^2|<br />

    The remaining terms are replaced by O(\Delta x^2) because of the existance of the bound that this implies, and that is all thet is needed for the rest of the argument. That is it saves us writing out a lot of detailed stuff that is not needed.

    CB
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  3. #3
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    Thanks for the reply! But it still doesn't answer my second question (or maybe I'm missing something ).
    Why is the function \Delta x^2 chosen in the first place? Why not \Delta x^3? I mean, how is the statement (in your reply) deduced?
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  4. #4
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    Quote Originally Posted by fishcake View Post
    Thanks for the reply! But it still doesn't answer my second question (or maybe I'm missing something ).
    Why is the function \Delta x^2 chosen in the first place? Why not \Delta x^3? I mean, how is the statement (in your reply) deduced?
    By looking at the form of the omitted terms, one of which is a multiple of \Delta x^2 which for small \Delta x dominates the other omitted terms.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    By looking at the form of the omitted terms, one of which is a multiple of \Delta x^2 which for small \Delta x dominates the other omitted terms.

    CB
    Ah, now I understand. I got it the other way round, since I was considering for large \Delta x and was wondering why it's not \Delta x^n. Thanks for the help!
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  6. #6
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    Quote Originally Posted by fishcake View Post
    Ah, now I understand. I got it the other way round, since I was considering for large \Delta x and was wondering why it's not \Delta x^n. Thanks for the help!
    By the way all of your limits should be \Delta x \to 0 not x \to \Delta x

    CB
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