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Binomial Theorem in Calculus

  1. Dec 24, 2008 #1
    I've started listening to the lectures for the MIT OpenCourseWare 18.01 Single Variable Calculus class. I understood all of it up until the teacher found the derivative of xn. Here's what he wrote on the board:

    [tex]\frac{d}{dx} x^{n} = \frac{\Delta f}{\Delta x} = \frac{(x+\Delta x)^{n} - x^{n}}{\Delta x}[/tex]

    That, I understand. Then we get to the binomial theorem to try to simplify [tex](x+\Delta x)^{n}[/tex]. The professor said that [tex](x + \Delta x)[/tex] is multiplied by itself n times, which I understand. Then he wrote:

    [tex](x+\Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + junk[/tex]
    [tex]junk = O((\Delta x)^{2})[/tex] ("big O of delta x squared")

    What I'm confused about is the last line. Why is that what "junk" equals? I understand that the rest of the terms don't matter as [tex]\Delta x[/tex] approaches 0, but why are they equal to what he says they are equal to "big O of delta x squared," as he said?
    Last edited: Dec 24, 2008
  2. jcsd
  3. Dec 24, 2008 #2


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    Big O

    Hi techieadmin! :smile:

    "= O((∆x)2)" is shorthand for "is of the order of (∆x)2" …

    for more details, see http://en.wikipedia.org/wiki/Big_O_notation :smile:
  4. Dec 24, 2008 #3


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    Well, it's exactly what your title implies- the binomial theorem.

    [tex](x+ y)^n= \sum_{i=0} \left(\begin{array}{c}n \\ i\end{array}\right)x^{n-i}y^i[/tex]
    When i= 0, [itex]\left(\begin{array}{c}n \\ 0\end{array}\right)[/itex] is 1 so the i=0 term is [itex]x^n[/itex]. When i= 1, [itex]\left(\begin{array}{c}n \\ 1\end{array}\right)[/itex] is n so the i= 1 term is [itex]nx^{n-1}y[/itex]. So
    [tex](x+ y)^n= x^n+ n x^{n-1}y+ \sum_{i= 2}^n\left(\begin{array}{c}n \\ i\end{array}\right)x^{n-i}y^i[/tex]
    It is that last sum that is the "junk" referred to.

    f(x)= O(g(x)) means that f(x) and g(x) go to the same limit "at about the same rate" as x goes to some value- specifically, that f(x)/g(x) has a no-zero finite limit.

    Here, the "junk" involves powers of [itex]\Delta x[/itex] of degree 2 and higher: you could write it as
    [tex]\left(\begin{array}{c}n \\ 2\end{array}\right)x^{n-2}(\Delta x)^2+ "other junk"[/tex]
    where the "other junk" are the remaining terms: involving powers of [itex]\Delta x[/itex] of degree 3 and higher. Dividing that by [itex](\Delta x)^2[/itex] will give [itex]\left(\begin{array}{c}n \\ i\end{array}\right)x^{n-2}[/itex] plus terms that still involve [itex]\Delta x[/itex] which will go to 0 as [itex]\Delta x[/itex] goes to 0 (which is, of course, the important value for a derivative).
  5. Dec 24, 2008 #4
    So the "junk" eventually approaches some number which is multiplied by [tex]\Delta x^{2}[/tex] and since [tex]\Delta x[/tex] goes to 0, the "junk" eventually disappears?
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