Proove logarithmic differentiation by induction

[Warr]

{\frac{P'(z)}{P(z)} = \frac{1}{z-z_1} + \frac{1}{z-z_2} + . . . + \frac{1}{z-z_n}

where P(z)=(z-z_1)(z-z_2)...(z-z_n)

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 \frac{1}{z-z_{k+1}} 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.

 

[VietDao29]

Hint: Have you tried differentiating P(z) using the product rule?
Something like:
(f(x) g(x) h(x))' = f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x).
Can you go from here? :)

 

[Warr]

yup I did that..

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

I mean I can get

\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)}

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.

 

[VietDao29]

Okay, so let:
P_x(z) = \prod_{i = 1} ^ x (z - z_i).
You need to prove:
\frac{P'_n(z)}{P_n(z)} = \sum_{i = 1} ^ n \frac{1}{z - z_i} (for n >= 1).
Of course:
\frac{P'_1(z)}{P_1(z)} = \sum_{i = 1} ^ 1 \frac{1}{z - z_i} = \frac{1}{z - z_1} is true.
Now, assume that:
\frac{P'_k(z)}{P_k(z)} = \sum_{i = 1} ^ k \frac{1}{z - z_i}
You need to prove that:
\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}}.
Now, note that: P_{k + 1}(z) = P_{k} (z) \times (z - z_{k + 1}). Can you differentiate P_{k + 1}(z) with repect to z in terms of P_{k}(z), \quad \mbox{and} \quad (z - z_{k + 1})?
From there, can you solve the problem?

 

[Warr]

ah yes, thanks :)