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I've been given the following equation;

$u''-\frac{f(z)}{z^2}u=0$ , with $f$ an analytic function:$\ \ $ $f(z) = f_0 + f_1 z + f_2 z^2 + \dots$

Using a substitution, of $u(z) = v(z)z^{\alpha}$ you can reduce the equation to:

$$v'' + \frac{2 \alpha}{z}v'-\frac{f_1 + f_2 z + f_3 z^2 + \dots}{z}v = 0$$

Then assuming $v(z) = 1 + v_1 z + v_2 z^2 + \dots$

I have found a formula for finding the coefficient $v_s$ by comparing each coefficient of $z^m$ in the above DE:

$$s(2\alpha + s - 1)v_s = \sum_{j=1}^s f_j v_{s-j}$$

I am then asked to find K and $\rho$ such that:

$$|f_s|\leq K \rho^{-s}$$

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