# Pade approximation of $\frac{1-e^{-x}I_0(x)}{x}$

I need to expand $\frac{1-e^{-x}I_0(x)}{x}$ in Pade approximation. The answer should be $\frac{1}{1+x}$. But I'm not sure how to reach the answer. Here $I_0(x)$ is modified Bessel function of order 0 and is given by $$I_0(x)=\sum_{s=0}^\infty \frac{1}{{s!}^2}\left(\frac{x}{2}\right)^{2s}$$ I looked for Pade approximation on online and found how to calculate it (set two polynomial on both denominator and numerator and compare with original function's Mclaurin expansion) but when I work it in polynomial of 1st order in numerator and 2nd order in denominator, I got $\frac{1}{1+\frac{3}{4}x}$ because original function is expanded like $1-\frac{3}{4}x+...$ , so I'm confused. Maybe the answer would be something like a round-off approximation. But I cannot justify it, because I cannot see the reason for giving up $\frac{3}{4}$ for $1$. But when I plot them, namely $\frac{1-e^{-x}I_0(x)}{x}, \frac{1}{1+x}, \frac{1}{1+\frac{3}{4}x}$, weirdly former one fits very well with original function. I have no idea how it can be possible. Can somebody help me?

• I hope that you noticed that the lazy way allows to build any Padé approximant. Commented Apr 17, 2018 at 8:25
• @Claude Leibovici Yes I noticed that but the given answer is not (0,1)order pade appriximation, though it fits better than it. So there is my confusion. Commented Apr 17, 2018 at 20:24
• I did misread since I thought you wanted the (1,2). For the (0,1), it is effectively $\frac{1}{1+\frac{3}{4}x}$. Sorry for the confusion. Cheers. Commented Apr 18, 2018 at 3:39

A simple way to do it is to consider the Taylor series first $$I_0(x)=1+\frac{x^2}{4}+\frac{x^4}{64}+O\left(x^6\right)$$ giving $$f(x)=\frac{1-e^{-x} I_0(x)}{x}=1-\frac{3 x}{4}+\frac{5 x^2}{12}-\frac{35 x^3}{192}+\frac{21 x^4}{320}+O\left(x^5\right)$$ which gives you the derivatives at $x=0$ $$f(0)=1 \qquad f'(0)=-\frac{3}{4} \qquad f''(0)=\frac{5}{6}\qquad f'''(0)=-\frac{35}{32}$$
Now, a $[1,2]$ Padé approximant is given by $$\frac {a_0+a_1 x}{1+b_1x+b_2 x^2}$$ where $a_0=f(0)$ and $$a_1=\frac{f(0)^2 f'''(0)+6 f'(0)^3-6 f(0) f'(0) f''(0)}{3 \left(2 f'(0)^2-f(0) f''(0)\right)}$$ $$b_1=\frac{f(0) f'''(0)-3 f'(0) f''(0)}{3 \left(2 f'(0)^2-f(0) f''(0)\right)}$$ $$b_2=-\frac{2 f'''(0) f'(0)-3 f''(0)^2}{6 \left(2 f'(0)^2-f(0) f''(0)\right)}$$
from which you would get $$f(x)=\frac{1+\frac{1}{7}x}{1+\frac{25 }{28}x+\frac{85 }{336}x^2}$$
If you are lazy (be sure that this is not a sin !), just write $$(1+b_1x+b_2x^2)f(x)=a_0+a_1x$$ and replace $f(x)$ by its Taylor series, expand and group terms. This would give $$(1-a_0)+(b_1-a_1-\frac 34)x+(b_2-\frac 34 b_1+\frac 5{12})x^2+\frac{1}{192} (80 b_1-144 b_2-35)x^3=0$$ Then, setting all coefficients equal to $0$ gives four linear equations in $a_0,a_1,b_1,b_2$ and the result.