Fractional Approximation of e^x

Question

I am trying to obtain an approximation for $e^x$ of the form $\frac{ax+b}{cx+d}$ using the Maclaurin series for $e^x$.

$\frac{ax+b}{cx+d}=e^x=1+x+\frac{x^2}{2}+\cdots\approx 1+x \\ ax+b = (1+x)(cx+d)=d+x(c+d)+cx^2\approx d+x(c+d) \\ 0 = (d-b) + x(c + d - a)$

which gives rise to two constraints: $d=b$ and $c+d-a=0\Rightarrow d = a-c=b$, which after substituting back into the original expression produces $$ e^x\approx \frac{ax+a - c}{cx + a - c}=\frac{a(x+1)+c}{c(x-1)-a} $$

Which reduces the expression down to 2 variables however I have no idea how to solve for the coefficients.

Answer 1

When $x\to 0$ then let's expand your fraction

$$\frac{ax+b}{cx+d}=\frac bd+\frac{(ad-bc)}{d^2}\ x+\frac{-(ad-bc)c}{d^3}\ x^2+ \frac{(ad-bc)c^2}{d^4}x^3+o(x^3)$$

And you have to identify to the Taylor Expansion of $e^x$.

• $\displaystyle 1=\frac bd\iff b=d$

• $\displaystyle 1=\frac{(ad-bc)}{d^2}\iff a-c=d$

• $\displaystyle \frac 12=\frac{-(ad-bc)c}{d^3}\iff d=-2c$

And we are done, since this completely determine $a,b,c,d$ (ignoring proportionality factor which will disappear in the fraction).

Note that the next term evaluates to $\displaystyle\frac{x^3}4$ so the approximation cannot go be better than $O(x^3)$.

Finally you get

$$\frac{2+x}{2-x}$$

Answer 2

In fact, what you want is to build the simplest Padé approximant of your function.

In a general manner, they write $$f(x)=\frac{\sum_{i=0}^n a_ix^i }{1+\sum_{i=1}^m b_ix^i }$$ Built around $x=a$, the simplest one write

$$f(x)=\frac{f(a)+ \left(f'(a)-\frac{f(a) f''(a)}{2 f'(a)}\right)(x-a) } { 1-\frac{ f''(a)}{2 f'(a)}(x-a)}$$