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Math Help - Negative Index & Binomial Theorem

  1. #1
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    Negative Index & Binomial Theorem

    hey everyone, I'm teaching myself calculus, because I felt it would be useful, even if I'm not going to use it in accounting. I am learning about the binomial theorem and differentiating negative indexes.

    Here is the current problem I'm working on
    1.) y=x^-2
    2.) y+dy=(x+dx)^-2
    3.) =x^-2(1+ dx/x)^-2
    4.) =x^-2 [1-2dx/x + 2(2-1)/1*2 * (dx/x)^2 - etc. ]
    5.) =x^-2 -2x^-3*dx + 3x^-4(dx)^2 -4x^-5(dx)^3 +etc...
    Ignoring small units
    6.) y+dy=x^-2 -2x^-3*dx
    Subtracting original y=x^-2
    7.) dy= -2x^-3*dx
    8.) dy/dx= -2x^-3

    the binomial theorem as it is explained in the book follows,

    (a+b)^n=a^n + n(a^n-1b/1) + n(n-1)(a^n-2b^2/2) + n(n-1)(n-2)(a^n-3b/3) + etc....

    I understand how one moves from Step 1 to 2, that being said how does (x+dx)^-2 become x^-2(1+dx/x)^-2? Secondly, could someone how one goes about applying the binomial theorem to a negative index?

    any help would be greatly appreciated
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  2. #2
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    Quote Originally Posted by Mausoldj View Post
    ....

    I understand how one moves from Step 1 to 2, that being said how does (x+dx)^-2 become x^-2(1+dx/x)^-2? Secondly, could someone how one goes about applying the binomial theorem to a negative index?

    any help would be greatly appreciated
    (x+dx)^{-2} = \left(x \cdot \left(1+\frac{dx}x \right) \right)^{-2} = x^{-2} \cdot \left(1+\frac{dx}x \right)^{-2}

    For example:

    (a+b)^{-2} = \frac1{(a+b)^2} = \frac1{a^2+2ab+b^2}
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  3. #3
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    Following the rule wouldn't it be,

    1.) (x+dx)'2
    2.) 1/(x'2) + 1/(2x*dx) + (dx/1)'2
    minus insignificant value
    3.) 1/x'2 + 2x*dx

    In your equation (x*(1+dx/x))'-2 shouldn't it be 1 over x? The rule you used doesn't seem to work for the problem.
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  4. #4
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    Quote Originally Posted by Mausoldj View Post
    Following the rule wouldn't it be,

    1.) (x+dx)'2
    2.) 1/(x'2) + 1/(2x*dx) + (dx/1)'2
    minus insignificant value
    3.) 1/x'2 + 2x*dx

    In your equation (x*(1+dx/x))'-2 shouldn't it be 1 over x? The rule you used doesn't seem to work for the problem.
    You can continue the first example using the second example:

    (x+dx)^{-2} = \left(x \cdot \left(1+\frac{dx}x \right) \right)^{-2} = x^{-2} \cdot \left(1+\frac{dx}x \right)^{-2} ~ = ~

     \frac1{x^2} \cdot \dfrac1{\left(1+\dfrac{dx}x \right)^{2}} = \frac1{x^2} \cdot \dfrac1{1+\dfrac{2dx}x + \dfrac{(dx)^2}{x^2}}

    But you can't split the last fraction into three different fractions with a single summand as denominator. Remember: The line of a fraction sets automatically parantheses - even though you can't see them.
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