If $y = e^{x(y-1)}$ then $y \approx 1-2(x-1)$ when $0<x-1<<1$

Question

Assume $y = e^{x(y-1)}$.

Then $y \approx 1-2(x-1)$ when $0<x-1<<1$

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I thought of something like that:

$$ e^{x(y-1)} = e^{-2(x-1)}e^{xy+x-2}=(1-2(x-1)+O(x-1)^2)e^{xy+x-2}$$

But I failed to prove that $e^{xy+x-2} \approx 1$

Any ideas?

Thanks

Answer 1

This isn't rigorous, but will give you a starting point: since $e^z \approx 1 + z + \dfrac 12 z^2$ you have $y = e^{x(y-1)} \approx 1 +x(y-1) + \dfrac 12 x^2 (y-1)^2$. That is, $$y-1 \approx x(y-1) + \frac 12 x^2 (y-1)^2$$ so that (after canceling the $y-1$) $$1 \approx x + \dfrac 12 x^2 (y-1)$$ and $$y \approx \frac{2}{x^2} (1-x) + 1$$ Since $0 < x-1 << 1$ you can approximate $\dfrac 2{x^2}$ by $2$.

Answer 2

Put $x:=1+u$, $\> y:=1+v$. Then your equation amounts to $1+v=e^{(1+u)v}$, or $$u={\log(1+v)\over v}-1=-{v\over2}+{v^2\over3}-{v^3\over 4}+\ldots\quad .$$ In order to obtain $v$ in terms of $u$ we have to invert the series on the right hand. Computing to order $5$ Mathematica produces $$v=-2u+{8\over3}u^2-{28\over9}u^3+{464\over135}u^4-{1496\over405}u^5 +?u^6\ .$$ From this we get $$y=1-2(x-1)+{8\over3}(x-1)^2-{28\over9}(x-1)^3+{464\over135}(x-1)^4-{1496\over405}(x-1)^5 +?(x-1)^6\ ,$$ which is valid for $|x-1|\ll1$.