Question regarding Langevin function and modified Bessel functions

Question

Upon noticing that the plots of the functions $f(x) = \frac{I_1(x)-I_1(-x)}{I_0(x)+I_0(-x)}$ and $L(x) = \coth(x)-\frac1{x}$ are nearly the same I wanted to ask whether the functions are just generally equal or at least approximately the same for small values of $x$ ($I_{\nu}(x)$ denotes the modified Bessel function). Thanks in advance!

Answer 1

First of all, notice that $$f(x)=\frac{I_1(x)}{I_0(x)}$$ Expanded around $x=0$ $$f(x)=\frac{x}{2}-\frac{x^3}{16}+\frac{x^5}{96}+O\left(x^7\right)$$ while $$L(x) = \coth(x)-\frac1{x}=\frac{x}{3}-\frac{x^3}{45}+\frac{2 x^5}{945}+O\left(x^7\right)$$ So $$\frac 3 2L(x)=\frac{x}{2}-\frac{x^3}{30}+\frac{x^5}{315}+O\left(x^7\right)$$ $$f(x)-\frac 3 2L(x)=-\frac{7 x^3}{240}+\frac{73 x^5}{10080}+O\left(x^7\right)$$ Now, they are much closer; for $x=1$, the difference is $-0.023$ while, before introducing the correction factor, it was $0.133$.