Approximation of $x\log x$

Question

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.)

Answer 1

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.

Answer 2

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 $\epsilon>0$.

Answer 3

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.