Desirae Washington

2022-07-02

Second Order Approximation for a Polynomial

if I have an expression: $L=\frac{12{a}^{3}{d}^{3}-4w{a}^{3}{d}^{2}+16{a}^{2}{d}^{2}-4w{a}^{2}d+6ad+1}{12{a}^{3}{d}^{3}-4w{a}^{3}{d}^{2}-4{a}^{2}wd+16{a}^{2}{d}^{2}+7ad-aw+1}$ what is the second order approximation in $\frac{d}{w}$?

I know that $(\frac{d}{w}{)}^{2}$ can be ignored but what about $\frac{{d}^{2}}{{w}^{3}}$. At this instant (without knowing the actual values of d wrt w) can we ignore this too? What about if we have (d/w=0.001)? Also how would the first order approximation in (d/w) be different in both cases?

if I have an expression: $L=\frac{12{a}^{3}{d}^{3}-4w{a}^{3}{d}^{2}+16{a}^{2}{d}^{2}-4w{a}^{2}d+6ad+1}{12{a}^{3}{d}^{3}-4w{a}^{3}{d}^{2}-4{a}^{2}wd+16{a}^{2}{d}^{2}+7ad-aw+1}$ what is the second order approximation in $\frac{d}{w}$?

I know that $(\frac{d}{w}{)}^{2}$ can be ignored but what about $\frac{{d}^{2}}{{w}^{3}}$. At this instant (without knowing the actual values of d wrt w) can we ignore this too? What about if we have (d/w=0.001)? Also how would the first order approximation in (d/w) be different in both cases?

jugf5

Beginner2022-07-03Added 18 answers

If the expression's numerator and denominator were both homogenous in $d$ and $w$ (for example, $\frac{d+w}{{d}^{2}-{w}^{2}}$ ) then it would be meaningful to ask for an n-th order expansion in $\frac{d}{w}$. The expression given is not homogenous in this sense, so approximation in $\frac{d}{w}$ without further specification (such as $\frac{d}{w}\ll 1;d\gg 1$ is not a well-posed question.

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