Approximations Involving Exponential Functions I am reading a text and I am curious to know how certain approximations were reached.
The first function approximations is: $$ 1- \frac{1}{2p}((1+p)e^{\frac{-y}{x(1+p)}} - (1-p)e^{\frac{-y}{x(1-p)}}) \approx \frac{y^2}{2x^2 (1-p^2)}$$
,when $y \ll x$. Note that I tried using the approximation $e^x \approx 1+x$, when x is small, but all I got was the conclusion that $1- \frac{1}{2p}((1+p)e^{\frac{-y}{x(1+p)}} - (1-p)e^{\frac{-y}{x(1-p)}}) \approx 0$.
The second function approximation is: $$ 1-e^{\frac{-y}{x}}(1-Q(a,b)+Q(b,a)) \approx \frac{y^2}{x^2 (1-p^2)}$$
,when $y \ll x$, where $Q(a,b) = \int_b^\infty e^{-\frac12 (a^2 + u^2)} I_0(au) u \, du$, $b = \sqrt{\frac{2y}{x(1-p^2)}}$, $a = bp$, $I_0$ is a modified Bessel Function of the first kind.
It is also a known fact that $$ Q(b,0) = 1$$ and $$ Q(0,b) = e^{-\frac{b^2}{2}}$$ I have tried to assume $$ a = 0$$, since $a = bp$  and $b$ is small, $ p$ is a number between 0 and 1. However, it is not clear how $(1-p^2)$ is in the denominator and not $(1-p^2)^2 $, which would be closer to the traditional $e^x$ approximation.
Any hints on how these approximations were derived would be appreciated. Thanks.
 A: For the first one, you need to keep one more term in the expansion. $e^x \approx 1+x+\frac{x^2}{2!}$.  When the first terms cancel, it is time for one more.  That is how the squares appeared.
A: For the second one, use the following approximations, valid when a,b,x are small: $$ Q(a,b) \approx 1 - \frac{b^2}{2} + \frac{a^2 b^2}{4} + \frac{b^4}{8}$$
$$ e^x \approx 1 + x + \frac{x^2}{2!}$$
Then after the substitution has been made, the result will be: $$1 - (A)$$
A has 15 terms, since it is the result of multiplying the 3 terms from the exponential approximation by (1+4+4 = 9) minus (2 from Q terms cancelling) minus (2 from 1, -1 cancelling). As an example of 2 of the 15 terms in A, $\frac{-y}{x} \frac{p^2}{(1-p^2)}$ and a higher term one: $\frac{-y^4}{4x^4 (1-p^2)^2}$.
Then after all 15 terms in A have been reached, eliminate all the terms containing $y^3$. I believe the reason is since y is much smaller than x, ($\frac{y}{x})^3$ is even smaller. Terms will cancel out and the desired approximation will be reached.
For reference, the approximation is defined on page 473, equation (10-10-10) of "Communication Systems and Techniques" by Mischa Schwartz.
