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I Partial Fraction Challenge

  1. Feb 17, 2017 #1
    I have figured out a nice way to prove that if the complex numbers [itex]z_1,z_2,\ldots, z_N\in\mathbb{C}[/itex] are all distinct, then the equation

    \prod_{n=1}^N \frac{1}{z - z_n} = \sum_{n=1}^N \frac{\alpha_n}{z-z_n}

    is true for all [itex]z\in\mathbb{C}\setminus\{z_1,z_2,\ldots, z_N\}[/itex], where the alpha coefficients have been defined by the formula

    \alpha_n = \underset{n'\neq n}{\prod_{n'=1}^N} \frac{1}{z_n - z_{n'}}

    I would like to leave the proof of this result as challenge to you guys, and I'm not in a need for advice myself at this point. Of course if somebody proves this in a way that is different from my proof, then I'm still reading.
  2. jcsd
  3. Feb 19, 2017 #2
    [ SPOILER="spoiler" ]

    There might be a simpler way, but you can prove it by induction. The base case is [itex]N=1[/itex]. Then we need to show that

    [itex]\Pi_{n=1}^1 \frac{1}{z-z_n} = \sum_{n=1}^1 \frac{\alpha_n}{z-z_n} [/itex]

    That's pretty obvious. Now, suppose it is true for [itex]N[/itex]. Then we have:

    [itex]\Pi_{n=1}^N \frac{1}{z-z_n} = \sum_{n=1}^N \frac{\alpha_n}{z-z_n}[/itex]

    Now, multiply both sides by [itex]\frac{1}{z-z_{N+1}}[/itex]. On the left side, we have:

    [itex]\Pi_{n=1}^{N+1} \frac{1}{z-z_n}[/itex]

    On the right side, we have:

    [itex]\sum_{n=1}^N \frac{\alpha_n}{z-z_n} \frac{1}{z-z_{N+1}}[/itex]

    Now, we use the fact that:

    [itex]\frac{1}{z-z_n} - \frac{1}{z-z_{N+1}} = \frac{z_n -z_{N+1}}{(z-z_n) (z-z_{N+1})}[/itex]

    So [itex]\frac{1}{z_n - z_{N+1}} [\frac{1}{z-z_n} - \frac{1}{z-z_{N+1}}]= \frac{1}{(z-z_n) (z-z_{N+1})}[/itex]

    So we can rewrite the right-hand side as

    [itex]\sum_{n=1}^N \frac{\alpha_n}{z_n - z_{N+1}} [\frac{1}{z-z_n} - \frac{1}{z-z_{N+1}}][/itex]

    [itex]= [\sum_{n=1}^N \frac{\bar{\alpha_n}}{z - z_n}] + \frac{\bar{\alpha_{N+1}}}{z-z_{N-1}}[/itex]

    where [itex]\bar{\alpha_n} = \frac{\alpha_n}{z_n - z_{N+1}}[/itex] and where [itex]\bar{\alpha_{N+1}} = \sum_{n=1}^N -\frac{\alpha_n}{z_n - z_{N+1}}[/itex]

    So we can write the right-hand side as [itex]\sum_{n=1}^{N+1} \frac{\bar{\alpha_n}}{z - z_n}[/itex]

    So the only thing left is to prove that [itex]\bar{\alpha_n}[/itex] has the right form, which is beyond what I have the energy for right now.
    [ /SPOILER ]
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