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when x is small, for example <1, then the expression can be approximate by (from a book)

$$ g(x)= \frac{-x^2}{8}{\frac { \left( 1-1/12\,{x}^{2} \right) }{1-1/4\,{x}^{2}} }= \frac{-x^2}{8} \frac{1}{1-1/6{x}^{2}} $$ (1.1)

let $$ f(x)= {\frac { \left( 1-1/12\,{x}^{2} \right) }{1-1/4\,{x}^{2}} }$$ the expression $g(x)$ should be dimensionless, so there should be keep a $ {x}^{2}$ in the denom part.

I have known that we can do this using the taylor series, like $taylor(f(x),x)$ is $(1+{\frac {1}{6}}{x}^{2}+{\frac {1}{24}}{x}^{4}+O \left( {x}^{6} \right) ) $ but this not in denom, a trick is $taylor(1/f(x),x)$ ,which will be $(1-{\frac {1}{6}}{x}^{2}-{\frac {1}{72}}{x}^{4}+O \left( {x}^{6} \right) ) $

so we can get the approximate like the $$ g(x)= \frac{-x^2}{8}{\frac { 1 }{1-1/6\,{x}^{2}} }$$ could other person give another method more directly, because I think this is too tricky.

and the second problem is, why we can not just in the numer make $1-x^2/12=1$, and the denom part not changed. howerver this will get a expression $$ g(x)= \frac{-x^2}{8}{\frac { 1 }{1-1/4\,{x}^{2}} }$$ (1.2)

(1.2) is not good like (1.1), we can just take a taylor, then we can compare.

but from the first eye, $x^2/8*(1-x^2/12)$ have a term $x^4$ ,when x is small, this can be dropped. but the fact is cannot, why? Does this means,if $x is not <<1$, but only $<1$, then we can not to drop a term, but have to do a taylor siries.

drop a term is only when the $x->0$, but taylor could get better approximation. even the term is two order high than other terms

comment, these approximate come frome a acoustic problem for a duct absorption problems, the impedance have a part like $$\left[ 1-\dfrac {2} {\sqrt {-i}k}\dfrac {J_{1}\left( \sqrt {-j}k\right) } {J_0\left( \sqrt {-jk}\right) }\right] ^{-1}$$

take the taylor series for $J_1$ and $J_0$, then get the polynom, the do another approximate, do this approximate, we can take the low frequency and high frequency easy.

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  • $\begingroup$ This site uses MathJax to format formulars so the look like real math. There a a tutorial here MathJax basic tutorial and quick reference. As it stands I'm not 100% sure what your formulas are. $\endgroup$ – Warren Hill Mar 9 '15 at 14:55
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    $\begingroup$ It is hard to see the function. At least could you put parentheses or tell what the numerator and what is the denominator ? $\endgroup$ – Claude Leibovici Mar 9 '15 at 14:59
  • $\begingroup$ 1. To convert an equation into MathJAX, just enclose it in dollar signs, so use \$x>8\$ to show $x>8$. 2. For a fraction, use \frac{numerator}{denominator} to show $\frac{numerator}{denominator}$. 3. For exponents use {base}^{exponent} to show ${base}^{exponent}$. $\endgroup$ – ghosts_in_the_code Mar 9 '15 at 15:18
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$$\frac{1}{8}x^2\frac{1-\frac{1}{12}x^2}{1-\frac{1}{4}x^2}=\frac18x^2\frac1{(1-(1/4)x^2)(1-(1/12)x^2)^{-1}}\approx\frac18x^2\frac1{(1+(1/12-1/4)x^2)}=\frac18x^2\frac1{(1-(1/6)x^2)}$$ As: $$(1+x)^n=1+nx+n(n-1)x^2/2+...\approx 1+nx\tag{$x\ll1$}$$ $$(1+ax)(1+bx)=1+(a+b)x+abx^2\approx1+(a+b)x\tag{$x\ll1$}$$

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  • $\begingroup$ a good answer, first change form, then taylor, is more natural than take taylor of the $1/f(x)$, thanks! $\endgroup$ – alljoyland Mar 9 '15 at 16:30

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