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I am trying to approximate the following expression for values of $m$ close to 0 and for values large values of $Ne$ ($m$ is bounded between 0 and 1 and $Ne$ is a positive integer)

$$X=-\frac{(m-1)^2}{2 m^2 \text{Ne}-m^2-4 m \text{Ne}+2 m-1}$$

I tried

Normal[Series[-((-1 + m)^2/(-1 + 2 m - m^2 - 4 m Ne + 2 m^2 Ne)), {m, 
   0, 1}]]

and also

Normal[Series[-((-1 + m)^2/(-1 + 2 m - m^2 - 4 m Ne + 
      2 m^2 Ne)) /. {m -> a/Ne}, {a, 0, 1}]] /. {a -> m Ne}

and both returned

$$1-4 m Ne$$

However $1-4 m Ne$ is actually a very poor approximation. The approximation of this expression is a classic result in population genetics. The approximation I am trying to reach is $\frac{1}{1+4 m Ne}$. $\frac{1}{1+4 m Ne}$ is indeed a good approximation as seen visually

Manipulate[
 Plot[{1 - 4 m Ne, 1/(
   1 + 4 Ne m), -((-1 + m)^2/(-1 + 2 m - m^2 - 4 m Ne + 
     2 m^2 Ne))}, {m, 0, 0.1}, PlotRange -> {{0, 0.1} {0, 0.1}}, 
  PlotStyle -> {Green, Red, Black}], {Ne, 1*^3, 1*^4}]

enter image description here

red line is $\frac{1}{1+4 m Ne}$, green is $1-4 m Ne$ (that returns negative values for most of the range of $m$ values) and the black line is the function that I am trying to approximate.

Can you help me to approximate this expression?

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3
  • $\begingroup$ I gave a computational way to arrive at the result in my answer, but I'm curious, why not just type 1/(1 + 4 m Ne)? $\endgroup$
    – Michael E2
    Aug 17, 2016 at 0:39
  • $\begingroup$ because, I am trying to learn to make approximations. I did not know about Pade approximation technic. You say it is a "computational way", but it is not a numerical approximation (but an analytical one), right? $\endgroup$
    – Remi.b
    Aug 17, 2016 at 0:43
  • $\begingroup$ By "computational" I meant computed (analytically) from the function by Mathematica, instead of typing out the expression based on math knowledge that you personally have. $\endgroup$
    – Michael E2
    Aug 17, 2016 at 1:08

2 Answers 2

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Perhaps this?

PadeApproximant[-((-1 + m)^2/(-1 + 2 m - m^2 - 4 m Ne + 2 m^2 Ne)), {m, 0, {0, 1}}]
(*  1/(1 + 4 m Ne)  *)

Note: A Taylor series is a power series, so one cannot hope for a rational approximation in the form sought. Note also that 1 - 4 m Ne is the Taylor series approximant to the Padé approximant above.

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2
  • $\begingroup$ I did not know about Pade approximation technic. Thanks +1. I don't know how you made the decision about the orders in the numerator and denominator but I suppose I'll have to further my understanding of Pade approximation to understand that, right? $\endgroup$
    – Remi.b
    Aug 17, 2016 at 0:47
  • 1
    $\begingroup$ @Remi.b Since the degrees of m in the numerator/denominator of the desired solution are 0/1 respectively, {0, 1} seems like the best thing to try first, yes? Wikipedia gives a pretty good summary of Padé approximation. $\endgroup$
    – Michael E2
    Aug 17, 2016 at 1:10
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If you inverse the given function, and inverse it's Taylor approximation, you would achieve a better approximation. it seems that the extreme changes around zero makes it unsuitable for Taylor expansion.

1/Normal[Simplify@Series[1/-((-1 + m)^2/(-1 + 2 m - m^2 - 4 m Ne + 2 m^2 Ne)), {m, 0,1}]] 

    (* 1/(1+4 m Ne)*)

Here is the verification of first order and second order approximations derived by this technique:

Manipulate[
 Plot[{-((-1 + m)^2/(-1 + 2 m - m^2 - 4 m Ne + 2 m^2 Ne)), 1/(
   1 + 4 m Ne), 1/(1 + 4 m Ne + 6 m^2 Ne)}, {m, 0, 0.1}, 
  PlotRange -> {{0, 0.1} {0, 0.1}}, 
  PlotStyle -> {Green, Red, Black}], {Ne, 1*^3, 1*^4}]

enter image description here

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