Expansion of $e^x$

Question

Find constants $A_1, A_2, A_3, A_4$ that $e^x=\frac{1+A_1x+A_2x^2}{1+A_3x+A_4x^2}+o(x^2)$ for $x\rightarrow0$. I know that $e^x=1+x+\frac{x^2}{2}+o(x^2)$, so should I just pick $A_1=1,A_2=\frac{1}{2},A_3=0,A_4=0$? I guess it's too easy haha

Answer 1

To approxiomate $e^x$ by $P/Q$ to the second order near $x = 0$, where $P(0) = 1$, $Q(0) = 1$ match $P = Q e^x$. Setting $Q(x) = 1 + ax + bx^2$, and using $e^x = 1 + x + ½x^2 + o(x^2)$, then: $$\begin{align} P(x) &= (1 + ax + bx^2)(1 + x + ½x^2) + o(x^2)\\ &= 1 + (1 + a)x + (½ + a + b)x^2 + o(x^2). \end{align}$$ Thus $$e^x = \frac{1 + (1 + a)x + (½ + a + b)x^2}{1 + ax + bx^2} + o(x^2).$$

A similar exercise matching to $o(x^4)$ determines $a = -½$, $b = 1/12$, $1 + a = ½$ and $½ + a + b = 1/12$. Thus $$e^x = \frac{1 + ½x + x^2/12}{1 - ½x + x^2/12} + o(x^4).$$

Edit: We can go two orders higher: $$e^x = \frac{1 + ½x + x^2/10 + x^3/120}{1 - ½x + x^2/10 - x^3/120} + o(x^6).$$