Approximation of $x\log(x)$ when $0

Question

Is there any approximation of $x\log(x)$ when $0<x<1$?

Answer 1

Since $x\log(x)$ is $0$ at the bounds, we can think about $$x\log(x)\approx x(x-1)P_n(x)$$ where $P_n(x)$ would be a polynomial of degree $n$.

To obtain the coefficients, we should minimize $$\Phi=\int_0^1\left(x\log(x)-x(x-1)P_n(x)\right)^2\,dx$$ For example $$P_1(x)=-\frac{91 x}{60}+\frac{133}{60}\qquad (\Phi=\frac{1}{2250})$$ $$P_2(x)=\frac{49 x^2}{20}-\frac{119 x}{30}+\frac{329}{120}\qquad (\Phi=\frac{1}{9600})$$ $$P_3(x)=-\frac{341 x^3}{70}+\frac{683 x^2}{70}-\frac{101 x}{14}+\frac{661}{210}\qquad (\Phi=\frac{38}{1157625})$$ $$P_4(x)=\frac{6149 x^4}{560}-\frac{7513 x^3}{280}+\frac{13849 x^2}{560}-\frac{1569 x}{140}+\frac{1949}{560}\qquad (\Phi=\frac{17}{1354752})$$ do not seem to be too bad.

I do not think that continuing will be of much interest, since one extra degree only reduces $\Phi$ by a factor almost equal to $2$ but, yes, we can continue improving.

We can also think about rational approximations; this one is not too bad $$\frac{5 (x-1) x \left(5 x^3+37 x^2+37 x+5\right)}{6 \left(x^4+16 x^3+36 x^2+16 x+1\right)}$$ giving a maximum error of $\approx 0.008$.

Answer 2

Depends on the acceptable errors: $$x\ln x = (x-1) \sqrt{x}$$ has maximum error of about $0.06$ and $$x\ln x = \sqrt{\frac{3}{4}}(x-1) \sqrt{x}$$ a more balanced maximum error of about $0.04$

Answer 3

Hint

Put $x=1-y$ and use Taylor expansion.

then

$$x\ln \left(x\right)=x\ln \left(1-y\right)=$$

$$x\left(-y+\frac{y^2}{2}-\frac{y^3}{3}+.....\right)=$$

$$x\left(x-1+\frac{\left(1-x\right)^2}{2}-.....\right)$$