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I want to approximate the expression $$ \frac{a z +1/2 b z^2 + c z^3}{a + b z + d z^2}.$$ This should be approximately equal to $z - \frac{b}{2a}z^2.$ But no matter the approach, I do not get this answer. Anyone who sees what I should have done?

Edit: some users had a fair point, in this case $z$ is very small

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  • $\begingroup$ What context does this come from? In addition, this certainly isn't equal to $z-\frac{b}{2a}z^2$. $\endgroup$ – Toby Mak Sep 11 at 10:14
  • $\begingroup$ You should state the neigbourhood in which we do the approximation. Is this approximation supposed to be valid when $z\to 0$? $z\to+\infty$? Something else? $\endgroup$ – Stinking Bishop Sep 11 at 10:19
  • $\begingroup$ is $z$ a complex number? Also use brackets. $\endgroup$ – NoChance Sep 11 at 10:29
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$$\begin{array}{rcl}\dfrac{az + \dfrac{1}{2} bz^2 + cz^3}{a + bz + dz^2} &=& \dfrac{az + bz^2 +dz^3-\dfrac{1}{2}bz^2-(d-c)z^3}{a +bz+dz^2}\\ &=&\dfrac{az + bz^2 +dz^3}{a+bz+dz^2}-\dfrac{\dfrac{1}{2}bz^2 +(d-c)z^3}{a+bz+dz^2}\\ &=&z -\dfrac{\dfrac{1}2b\left(z^2 +\dfrac{2(d-c)}{b}z^3\right)}{a+bz+dz^2}\\ &=& z - \dfrac{\dfrac{b}{2a}\left(az^2 +\dfrac{2a(d-c)}{b}z^3\right)}{a+bz+dz^2}\\ &=& z - \dfrac{b}{2a}\cdot \dfrac{az^2+bz^3+dz^4-dz^4+\dfrac{2a(d-c)-b^2}{b}z^3}{a+bz+dz^2}\\ &=&z-\dfrac{b}{2a}z^2 +\dfrac{b}{2a}\cdot\dfrac{dz^4-\dfrac{2a(d-c)-b^2}{b}z^3}{a+bz+dz^2}\\ \end{array}$$

This should be approximately equal to $z-\dfrac{b}{2a}z^2$ when $z$ is sufficiently small.

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Why do you think this should be approximately equal to $z-\frac{b}{2a} z^2$?

$$\frac{z^2 \left(cz+1/2 \cdot b+a/z \right)}{z^2 \left(d+b/z+a/z^2 \right)}$$ $$= \frac{cz+1/2 \cdot b+a/z}{d+b/z+a/z^2}$$

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