# Approximation of $x\log x$

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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• 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? – David White Jun 1 at 1:12

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.

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 $$\epsilon>0$$.

• 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? – Jordi Jun 1 at 0:34

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.