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Thread: Polynomial division and dividing by zero

  1. #1
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    Polynomial division and dividing by zero

    Suppose I define $\displaystyle g(x,y) = \frac {f(x)-f(y)}{x-y} $ and take $\displaystyle f(z) = z^{n} $. So $\displaystyle g(x,y) = x^{n-1}+ x^{n-2}y + \cdots + xy^{n-2} + x^{k-1} $. If I take $\displaystyle y=x $, then $\displaystyle g(x,x) = nx^{n-1} $. Is this valid? If I look at the original equation, does this also mean $\displaystyle g(x,x) = \frac {0}{0} $? I'm not sure how to understand division by zero.
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  2. #2
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    $\displaystyle g(x,x) \neq nx^{n-1}$ because deriving $\displaystyle g(x,y) = x^{n-1}+ x^{n-2}y + \cdots + xy^{n-2} + x^{k-1}$ from $\displaystyle g(x,y) = \frac {f(x)-f(y)}{x-y}$ is valid only when $\displaystyle x \neq y$.

    Again, if $\displaystyle x=y$ then $\displaystyle g$ is not defined, therefore you can't say that $\displaystyle g(x,x) = nx^{n-1}$
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  3. #3
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    OK, the problem I'm doing says $\displaystyle g(x,y) $ is always divisible by $\displaystyle x - y $, so instead, can I say because $\displaystyle (x-y)g(x,x) = (x-y)(nx^{n-1}) $, and then using the cancellation law, $\displaystyle g(x,x)$ is the same polynomial as $\displaystyle nx^{n-1}$?
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