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I was wondering if anyone knows a good approximation for the function $$f(x)=x\log x $$ when $x$ goes to infinity. In particular, I would like to get rid of the $\log x$ and so I need a polynomial approximation. (I think Stirling would not work very well.)

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  • $\begingroup$ You have me curious. If you already have the function, why do you want to approximate it? Also, have you tried a Taylor or Maclaurin series expansion? $\endgroup$ – David White Jun 1 at 1:12
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Let $x=a+t(b-a)$ for some $t$ in the range $0$ to $1$. You get :

Let $c=b-a$

$$y = (a+ct)log(a+ct)$$

$$y=(a+ct)log\left( a\left(1+\frac c a t\right) \right)$$

$$y = (a+ct)log(a) + (a+ct)log\left(1+\frac c a t\right)$$

You can now replace the log() in the RHS by any suitable polynomial covering a reasonable range of values.

Note that using the arbitrary choice $a=c$ greatly simplifies things.

Good expressions for $log(1+x)$ are possibly better chosen from Páde approximants than from ordinary polynomials. This page particularly mentions $log(1+x)$ Páde expressions and nearer home this Q&A on Mathematics SE describes exactly this.

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There is no adequate polynomial or power law approximation. For large $x$, $f(x)=x\log x$ grows more slowly than $x^{1+\epsilon}$ and more rapidly than $x^{1-\epsilon}$ for every $\epsilon>0$.

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  • $\begingroup$ Hey Buzz, thanks for the comment. I understand what you mean, but then, should I just say that there is no good approximation? My problem comes from the fact that I have an equation in which such terms appears, and I need to find x as a function of the rest; so what could I do? Just say that it is not possible to solve it? $\endgroup$ – Jordi Jun 1 at 0:34
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By the substitution $u=\frac{1}{x}$ you can convert this to approximating $g(u)=-\displaystyle{\frac{\ln(u)}{u}}$ for $u\to0$. Then you can use the Taylor series for $g(u)$ centered at $u=1$, for example.

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