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I have some fractional expression like

$$ \frac{x^2+xyz+z^2}{x^2-yz+x^2z} $$

and I know that $x \ll y \ll z$. Really I want to divide through by $z^2$ and then take a Taylor series expansion in terms of $\frac{x}{z}$ and $\frac{y}{z}$.

I have tried this approach:

  1. Divide the numerator alone by $z^2$ using Numerator
  2. Replace terms like $\frac{x}{z} \to \frac{x}{z} \epsilon$
  3. Then I would take a Taylor series in $\epsilon$ and let $\epsilon \to 1$. (or maybe just set terms like $\epsilon^2 \to 0$)

I have two questions:

  1. Is there a better way than separating the numerator and denominator and treating them separately, and more pressingly

  2. ReplaceAll (/.) won't work on a compound expression like $\frac{x y}{z^2}$, I want $\frac{x y}{z^2} \to \frac{x y}{z^2} \epsilon^2$ but the replacement doesn't work on these compound expressions

Thanks

EDIT: sorry about the fraction above, it won't let me post between $$ $$ symbols

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  • $\begingroup$ Replace x and y by the same quantities multiplied by ϵ and use Series for the expansion in ϵ. $\endgroup$ – bbgodfrey Nov 23 '17 at 17:19
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With $x\ll z$ and $y\ll z$ I would substitute $x$ with $az$ and $y$ with $bz$. Under these conditions we have $a \approx 0$ and $b \approx 0$. Then take your Taylor expansion around $(a,b)=(0,0)$, then perform the reverse substitution $a$ with $x/z$ and $b$ with $y/z$.

f = (x*x + x*y*z + z*z)/(x*x - y*z + x*x*z)
orderN = 2

Simplify[
    Normal[
            Series[f/z^2 /. x -> a*z /. y -> b*z, 
                   {a, 0, orderN}, {b, 0, orderN}]
        ] /.  a -> x/z /. b -> y/z
 ]

which gives: $$ -\frac{x^2 \left(y+z^2+z\right)+x y^2 z+y z^2}{y^2 z^3} $$

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As per the comment, replacing $x \to x \epsilon$ and $y \to y \epsilon$ works in this simple case

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