Let us consider the first example in Ch. 15 of Hull: Options, Futures, and other derivatives (6th edition).

$S_0=49$, $K=50$, $r=0.05$, $\sigma=0.2$, $T=0.3846 = 20/52$.

The price of the option per share should come out to $2.4$ according to Hull. We will check this against the Black-Scholes pricing formula for a call option $c$ on page 295.

We first calculate by hand that $\sigma^2/2=1/50$, $r\pm \sigma^2/2=(5\pm 2)/100$, and

$(r\pm\sigma^2/2)T = \frac{5\pm 2}{(13)(20)}$.

Moreover $S_0/K = 49/50$. A good approximation to the natural logarithm for inputs near 1 is $\ln x \approx x-1$.

Thus $\ln (S_0/K)\approx -\frac{1}{50}$.

Hence $(r+\sigma^2/2)T + \ln(S_0/K) \approx \frac{5+2}{(13)(20)} – \frac{1}{50} = \frac{9}{1300}$. This is the numerator of $d_1$. The denominator is $\sigma\sqrt{T} = \sqrt{5/13}/5 = 1/\sqrt{65} \approx 1/\sqrt{64} = 1/8 = .125$. So we have

$d_1 \approx \frac{9/1300}{1/8} = \frac{18}{325}$. It is tempting to say this is approximately $6\%$; a little calculator help shows it is approximately $5.5\%$. Then we also have $d_2 = d_1-\sigma\sqrt{T}\approx 5.5\% – 12.5\% = -7\%$.

Next we need to approximate the cumulative distribution function of the standard normal distribution, $N(d)$. Note that $N(0)=1/2$ and that $N’(0)=e^{-0^2/2}/\sqrt{2\pi} = 1/\sqrt{2\pi}$, so a linear approximation for $d\approx 0$ is $N(d)\approx \frac{1}{2} + \frac{d}{\sqrt{2\pi}}$.

A quite good approximation for $\sqrt{2\pi}$ is 2.5 (a little calculator help helps here).

This gives $N(d_1)\approx \frac{1}{2} + \frac{11}{500}$ and $N(d_2)\approx \frac{1}{2} – \frac{14}{500}$.

Next we approximate the exponential function near zero: $e^{-rT}\approx 1-rT = 1-\frac{1}{52}$ and $Ke^{-rT}\approx 50(1-\frac{1}{52})$.

Finally, $c = S_0 N(d_1) – K e^{-rT} N(d_2) = \frac{1}{2}(-\frac{1}{26}) + \frac{1}{500}(49\cdot 11 + 50(\frac{51}{52})(14))$. To get a good approximation we replace $50(51/52)$ by $49$, and get

$c \approx \frac{1225}{500} – \frac{1}{52} \approx \frac{2450}{1000} – \frac{1}{50} = 2.43$ which agrees to the precision given with our desired answer $2.4$.