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Basic Derivation with negative exponents

  1. Jul 9, 2009 #1
    Hello reader

    So I'm trying to teach mysel calculus with Thompson's Calculus made easy.
    I'm still in the very early stages and I got stuck with negative exponents.
    I'd basically like an explanation fof what's going on here.

    http://www.flickr.com/photos/26906498@N03/3705112800/ [Broken]

    I've tried using online sources like:
    http://www.purplemath.com/modules/exponent.htm
    http://www.algebrahelp.com/lessons/simplifying/negativeexponents/

    I also looked up Binomial Theorem on wikipedia which just confused me more.
    http://en.wikipedia.org/wiki/Binomial_theorem
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 9, 2009 #2

    rock.freak667

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    Well basically, the binomial theorem is valid for only n being an integer such that

    [tex](a+b)^n = b^n +^nC_1 b^{n-1}a + ^nC_2 b^{n-2}a^2 +...+ ^nC_r b^{n-r}a^r + a^n[/tex]


    But if you use the definition of nCr being n!/(n-r)!r!. Then terms like nC1 works out as

    [tex]^nC_1 = \frac{n!}{(n-1)!1!} = \frac{n(n-1)!}{(n-1)!}=n[/tex]

    [tex]^nC_2 = \frac{n!}{(n-2)!1!} = \frac{n(n-1)(n-2)!}{(n-1)!2!}= \frac{n(n-1)}{2}[/tex]

    So (a+b)n simplifies to this

    [tex](a+b)^b = b^n + nb^{n-1}a + \frac{n(n-1)}{2!}b^{n-2}a^2 + \frac{n(n-1)(n-2)}{3!}b^{n-3}a^3 + ...[/tex]


    so if ever you put n as a negative number or a fractional number then you'd have an infinite number of terms which is valid for |b/a| < 1.

    so in the chapter, the expansion is valid for

    [tex]\left | \frac{dx}{x} \right | < 1 \Rightarrow -1< \frac{dx}{x}<1[/tex]

    or simply dx/x is small, that is why they neglected the terms after x-3


    EDIT: Normally you won't need to do the y+dy = x +dx thing always, you can just usually apply y=xn => dy/dx=nxn-1
     
  4. Jul 9, 2009 #3

    HallsofIvy

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    rockfreak667, if you go down far enough on the wikipedia page cited, you find "Newton's Generalized Binomial Formula" which does NOT require positive integers.

    For any non-negative integers, n and m, [itex]n\ge m[/itex]
    [tex]\left(\begin{array}{c}n \\ m\end{array}\right)= \frac{n!}{m!(n-m)!}= \frac{n(n-1)(n-2)\cdot\cdot\codt\(n-m+1)}{m(m-1)(m-2)\cdot\cdot\cdot(3)(2)(1)}[/tex]
    In that final form, we can, in fact, write it out for n and m any numbers, not just non-negative integers. Of course,
    [tex]\left(\begin{array}{c} n \\ n\end{array}\right)= 1[/itex]
    and
    [tex]\left(\begin{array}{c}n \\ n-1\end{array}\right)= n[/itex]
    for any n, non-negative integer or not. Thus, we can write
    (x+ h)n= (1)xn+ (n)x{n-1}h+ terms involving h2 or higher, for any number n, positive integer or not. Then, (x+h)n- xn= n xn-1h+ terms involving h2 or higher and, dividing by h, [(x+h)n- xn]/h= n xn-1+ terms involving h or powers of h. Taking the limit as h goes to 0, all those "terms involving h or powers of h" go to 0 leaving only nxn-1.
     
  5. Jul 9, 2009 #4

    rock.freak667

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    That's how I learned the binomial theorem, first with n being a positive integer and then the generalized form for fractions and negative numbers. Most likely due to working out 50C34 and such we just used a calculator instead of writing it out...things like -3C1 on a calculator would make me go :bugeye:
     
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