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Thread: Proof of Lemma on Polynomials

  1. #1
    Sep 2009

    Proof of Lemma on Polynomials

    There is a Lemma in my textbook that states,

    If P(h) is a polynomial of degree <= k, that vanishes to order > k as h -> 0 [i.e. P(h)/|h|^k -> 0], then P === 0 (=== is a triple equals sign)


    I don't quite understand what P===0 means. I know that the === represents congruence used in modular arithmetic. However, how is this applicable here?

    Also, I don't understand how P(h)/|h|^k -> 0 implies that polynomial of degree <= k vanishes to order > k. From this statement, it seems like P vanishes to order equal to k.

    Basically, what does this Lemma even mean?
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  2. #2
    Super Member
    Apr 2009
    If $\displaystyle P \in F[x]$ (where $\displaystyle F= \mathbb{R}$ or $\displaystyle \mathbb{C}$ ) $\displaystyle P(x)= \sum_{j=0} ^{k} \ a_jx^j$ and $\displaystyle \frac{P(x)}{x^k} = \sum_{j=0} ^{k} \ a_jx^{j-k}$ and so for every $\displaystyle i<k$ the term $\displaystyle x^{i-k} \rightarrow \infty$ as $\displaystyle x \rightarrow 0$ (assuming $\displaystyle a_i \neq 0$) and so for all $\displaystyle i<k$ $\displaystyle a_i=0$, and so $\displaystyle 0= \frac{P(x)}{x^k} = a_k$ and so$\displaystyle P \equiv 0$ (this means P is identically zero). Now try induction on the number of variables.
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