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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

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

    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

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

    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

    stevendaryl

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    Staff Emeritus
    Science Advisor

    [ 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 ]
     
  4. Mar 21, 2017 #3
    A month has passed since the opening post, so I'll give a hint. My own proof splits into two independent parts, where we use one technique to prove that if the formula is true with at least some alphas, then the alphas cannot be anything else but the ones given by the claim. Then we use another technique to prove that the formula is true with at least some alphas. In other words my proof proves the uniqueness and existence of alpha coefficients separately, which is typical in mathematics.

    Although the work of stevendaryl was left incomplete, actually the induction step found by him does provide a proof for the existence of the alphas, so in this sense half of the problem already got solved. Anyway, I can assure you that my proof for the existence does not use induction, and consequently in my opinion has more style to it. I cannot insist that my existence proof would be shorter though; looks roughly the same in terms of length.
     
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