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So i have the following relation: $$x\cdot g(y(x)) = 1$$ Taking the logarithm of this equation yields $$\log x + \log g(y) = 0 $$

I work with the following assumptions: $(x,y(x)) = (1,0)$ satisfies the relation, $g(0)=1$ and

$$\frac{d}{dy}[\log g(y)]_{y=0} = - a$$ $$\frac{d^2}{dy^2}[\log g(y)]_{y=0} = b $$

It doesnt matter what $a,b>0$ are, but supposedly $y(x)$ admits the following expansion in $\log x$ provided that $\log x$ is close to 0

$$y(x)= \frac{1}{a}\cdot \log x + \frac{1}{2}\cdot \frac{b}{a^3} (\log x)^2 + \mathcal{O}(\log x)^3$$

I just dont understand how to do this... If i take derivatives w.r.t. to $x$, using that $y(1)=0$ i find $y'(1)= \frac{1}{a}, y''(1)= \frac{b}{a^3}-\frac{1}{a}$ which would be the coefficients of the normal Taylor expansion. (these very much look like the coefficients in the given expansion)

Im just puzzled. I dont see how to make such expansion at all..

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