Estimate an exponential and expand a shifted secant

$\ln f=\tan^{-1}(x/2)$ gives $f^{\prime}/f=(1/2)/(1+x^2/4)=2/(4+x^2)$, hence $(4+x^2)f^{\prime}=2f$.

$(4+x^2)f^{\prime\prime}+2xf^{\prime}=2f^{\prime}$, so $(4+x^2)f^{\prime\prime}=(2-2x)f^{\prime}$. Thus $f(0)=1$, $f^{\prime}(0)=1/2$, $f^{\prime\prime}(0)=1/4$, and $f(x)=1+x/2+x^2/8+O(x^3)$.

Choose $x=2/\sqrt3$, so $\tan^{-1}(x/2)=\pi/6$. The estimate is $1+1/\sqrt3+1/6=7/6+1/\sqrt3\approx1.74401694$, printed as $1.74402$ or $1.74$ (3 s.f.). The relative error is $100|(7/6+1/\sqrt3)-e^{\pi/6}|/e^{\pi/6}\approx3.313\%$, about $3.3\%$. Calculate this before rounding the estimate to $1.74$.

Set $u=x/a$. Since $\sqrt2\cos(u+\pi/4)=\cos u-\sin u=1-u-u^2/2+O(u^3)$,

$\begin{aligned}&g(x)\\ & =\left[1-\left(u+\frac{u^2}{2}+O(u^3)\right)\right]^{-1}\\ & =1+u+\frac32u^2+O(u^3).\end{aligned}$

Hence $p=1/a$ and $q=3/(2a^2)$. The square root applies only to the factor $2$, not to the whole denominator; this is confirmed by the printed angle-addition working.