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I would like to find approximative (polynomial) expression for $x^2 ln(x)$ in the vicinity of x=0, but any expansion (e.g. the last row of the first table on http://math2.org/math/expansion/log.htm) seems to diverge. On the other hand, the x->0 limit of the above expression, by using the L’Hopital’s Rule, is 0. Does anyone has any idea how to get the approximative expression?

The reason why I am looking for an approximation is because I want to find the simplest possible expression, i.e., dependence on x. The full expression reads $\frac{1 -x^2 ln(2/x)}{1+x^2}$, and by plotting it in the range (0.1,0.7), I found (fit, or Mean Value Theorem, but not analytically) that sufficiently good approximation (within few percent) is $1-x$. But, I also need the same expression to be valid for my entire $x$ range, from 0.02 up to 0.7.

Boki
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  • Since $\lim_{x\to0^+}x^n\ln x=0$ for all $n\in\Bbb Z^\ast$, $\ln x$ isn't asymptotic to any polynomial or reciprocal thereof. What alternative kind of approximation would suit you depends on why you want to approximate $x^2\ln x$ in the first place, so please explain that in an edit. – J.G. Nov 16 '21 at 14:30
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    https://math.stackexchange.com/questions/1046321/approximating-log-x-with-roots – Miscellaneous Nov 16 '21 at 14:33
  • @Miscellaneous Thanks, but to me it seems that the approximation of ln(x) (mentioned in math.stackexchange.com/questions/1046321/… ) works better around x=1, i.e. for higher values, and I need it to be good at x close to zero. – Boki Nov 16 '21 at 15:37
  • @Boki Would a numerical solution be adequate for your needs? – njuffa Nov 18 '21 at 00:17
  • Have a look at the update. – Claude Leibovici Nov 22 '21 at 12:00

2 Answers2

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Since $$\lim_{x\to 0} \, x\log(x)=\lim_{x\to 0} \, (x^x-1)$$ you could use,as an approximation, $$x^2 \log(x) \sim x(x^x-1)$$

To give an idea, consider the norm $$\Phi=\int_{0}^{\frac 1{10}}\Big[x^2 \log(x) -x(x^x-1)\Big]^2\,dx=1.14\times 10^{-7}$$

Edit

Graphically, for $0 \leq x \leq 1$ $$x^2 \log(x) \sim (x-1) x \sqrt x$$ So, consider the norm $$\Phi(a)=\int_0^a \Big[x^2 \log(x) -k_a (x-1) x \sqrt x\Big]^2\,dx$$ which is explicit. Now, solve, for $k$, $\Phi'(a)=0$. This gives $$k_a=\frac{40 \sqrt{a} (-162 a+891 a \log (a)-1089 \log (a)+242)}{3267 \left(10 a^2-24a+15\right)}$$

If you want the best approximation for $a=\frac 1{100}$, $$k_{\frac 1{100}}=\frac{40 (24038+216018 \log (10))}{48224187}\sim \frac{298}{689}$$ and $$\Phi \left(\frac{1}{100}\right)=2.38 \times 10^{-12}$$

Update

We can do a bit better using the previous similarity $$x^2 \log(x) \sim (x-1) x \sqrt x$$ and use as an approximation $$x^2 \log(x) \sim \sum_{k=0}^p A_k^{(p)} x^{\frac{k+3}2}$$

This would give $$ A_0^{(1)}=\frac{8}{81} \sqrt{a} \,(9 \log (a)-2)\implies \Phi \left(\frac{1}{100}\right)=2.27 \times 10^{-12}$$ $$ A_0^{(2)}=-\frac{8 \sqrt{a}}{5} \qquad A_1^{(2)}=\log (a)+\frac{71}{45}\implies \Phi \left(\frac{1}{100}\right)=9.88 \times 10^{-15}$$

Claude Leibovici
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  • Thanks, that helped a lot. I am a bit rusty, and it was hard for me to come up with the idea that $x^x=e^{\ln{x^x}}$, and to expand the exponent. In the end, it turns out, that for my calculations $x^{x+1}$ could be neglected compared to $x$ in the vicinity of $x=0$. – Boki Nov 18 '21 at 18:18
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For some identities, it is fruitful to look at

log(x)x x

This is a simple example to apply the Taylor expansion. It is an example of the by far more general formulare:

$f(0)ln(x)+ln(x)f'(0)x+ln(x)/2f''(0)x^2+O(x^3)$

with

$f(x)=x^2$

In this example this is the identity to the third-order Taylor expansion is the same as the product function given up to the third-order correction term.

This result can be verified on the name computational knowledge engine from Wolfram. It is close to the Taylor expansion of function with the exponential function giving a similar polynomial for $f(x_0)$. Simply substitute $t=exp(x)$ to get the logarithm out of the expansion function product. It is then rather difficult to deal with the transformation of $0\rightarrow -\infty$. There is a complex function theory needed to prove a deal further with that. Put the expansion as given, known and valid maths is much easier to cope with.

This knowledge is astonishingly rare among Mathematicians. It is fairly common among Electronic Engineers in some of their disciplines. Despite it belong to complex function theory Taylor expansion methodology.

Hope the helped and kept the answer short enough. In short prove theory of informal extend the graphical match shall be prove enough, too enough since it is identic and therefore special.

Steffen Jaeschke
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  • In your example $f(0)=0$ and $f'(0)=0$, so your expansion just says $x^2ln(x) = 0 + 0 + x^2ln(x) + O(x^3)$. – quarague Nov 17 '21 at 09:19