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Homework Help: Proove logarithmic differentiation by induction

  1. Feb 2, 2006 #1
    [tex]{\frac{P'(z)}{P(z)} = \frac{1}{z-z_1} + \frac{1}{z-z_2} + . . . + \frac{1}{z-z_n}[/tex]

    where [tex]P(z)=(z-z_1)(z-z_2)...(z-z_n)[/tex]

    Any hints? I've shown it works for a few specific cases..now I have to show that it works for n=k+1. I tried adding a [tex]\frac{1}{z-z_{k+1}}[/tex] term to both sides, and trying the product rule for P'(z)..but couldn't really get anywhere.

    btw, these are complex functions, althought I don't think it makes a difference here.
     
  2. jcsd
  3. Feb 2, 2006 #2

    VietDao29

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    Hint: Have you tried differentiating P(z) using the product rule?
    Something like:
    [tex](f(x) g(x) h(x))' = f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x)[/tex].
    Can you go from here? :)
     
  4. Feb 2, 2006 #3
    yup I did that..

    I know I can directly show it..but is this induction?

    I mean I can get

    [tex]\frac{(z-z_2)...(z-z_n) + (z-z_1)(z-z_3)...(z-z_n) + ... (z-z_1)...(z-z_{n-1})}{(z-z_1)(z-z_2)...(z-z_n)}[/tex]

    which obviously gives the R.S of the eq'n in my first post, but is this 'Shown by induction on n'? I thought I had to interate by 1 term and then prove that it holds to the pattern.
     
  5. Feb 3, 2006 #4

    VietDao29

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    Okay, so let:
    [tex]P_x(z) = \prod_{i = 1} ^ x (z - z_i)[/tex].
    You need to prove:
    [tex]\frac{P'_n(z)}{P_n(z)} = \sum_{i = 1} ^ n \frac{1}{z - z_i}[/tex] (for n >= 1).
    Of course:
    [tex]\frac{P'_1(z)}{P_1(z)} = \sum_{i = 1} ^ 1 \frac{1}{z - z_i} = \frac{1}{z - z_1}[/tex] is true.
    Now, assume that:
    [tex]\frac{P'_k(z)}{P_k(z)} = \sum_{i = 1} ^ k \frac{1}{z - z_i}[/tex]
    You need to prove that:
    [tex]\frac{P'_{k + 1}(z)}{P_{k + 1}(z)} = \sum_{i = 1} ^ {k + 1} \frac{1}{z - z_i} = \sum_{i = 1} ^ {k} \left( \frac{1}{z - z_i} \right) + \frac{1}{1 + z_{k + 1}}[/tex].
    Now, note that: [tex]P_{k + 1}(z) = P_{k} (z) \times (z - z_{k + 1})[/tex]. Can you differentiate [tex]P_{k + 1}(z)[/tex] with repect to z in terms of [tex]P_{k}(z), \quad \mbox{and} \quad (z - z_{k + 1})[/tex]?
    From there, can you solve the problem?
     
    Last edited: Feb 3, 2006
  6. Feb 3, 2006 #5
    ah yes, thanks :)
     
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