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Show given expressions are polynomials by converting them to canonical form

Transform the given expression to the canonical form \sum_{k=0}^n c_k x^k to prove that they are, in fact, polynomials.

  1. (1+x)^{2n}.
  2. \displaystyle{\frac{1-x^{n+1}}{1-x}}.
  3. \prod_{k=0}^n \left(1+x^{2^k}\right).

  1. Proof. We use the binomial theorem,

        \begin{align*}  (1+x)^{2n} &= \sum_{k=0}^{2n} \binom{2n}{k} x^k \\ &= \sum_{k=0}^m \binom{m}{n} x^k & (\text{letting } m = 2n)  \\ &= \sum_{k=0}^m c_k x^k & (\text{letting } c_k = \binom{m}{k} \text{ for each } k). &\qquad \blacksquare \end{align*}

  2. Proof. Recalling a previous exercise, we have

        \begin{align*}  \frac{1-x^{n+1}}{1-x} &= \frac{(1-x)(1+x+ \cdots + x^n)}{1-x} \\ &= 1+x+ \cdots + x^n \\ &= \sum_{k=0}^n 1 \cdot x^k. &\blacksquare \end{align*}

  3. Proof. Again, we recall a previous exercise to obtain,

        \begin{align*}  \prod_{k=0}^n \left( 1+x^{2^k} \right) &= \frac{\left(1 - x^{2^{n+1}} \right)}{1-x} \\ &= \frac{\left(1-x^{2^n}\right)\left(1+x^{2^n}\right)}{1-x} \\ &= \left( \frac{1-x^{2^n}}{1-x} \right) \left( 1+x^{2^n} \right) \\ &= \left(1+x+\cdots + x^{2^n -1} \right) \left( 1+x^{2^n} \right) \\ &= \left( 1+x+ \cdots + x^{2^n -1} \right) \left( x^{2^n} + x^{2^n +1} + \cdots + x^{2^{n+1} -1} \right) &(\text{distributing}) \\ &= \sum_{k=0}^{2^{n+1}-1} 1 \cdot x^k \\ &= \sum_{k=0}^m 1 \cdot x^k & (m = 2^{n+1}-1). && \blacksquare \end{align*}

2 comments

  1. Jacare says:

    Hi. Is there an addition sign (+) missing in the 5th line in part c? The line that has “(distributing)” to the right of it.

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