2
$\begingroup$

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.)

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ Jun 1 '19 at 1:12
3
$\begingroup$

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 $\epsilon>0$.

$\endgroup$
1
  • $\begingroup$ 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 '19 at 0:34
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.