$n$th derivative of $\sqrt{\frac{(s_0+T)(s_0 + T + 2\sqrt{a})}{(s+T)(s+ T+2\sqrt{a})}}$

Question

I'm trying to obtain the $n$th derivative of

$$f(s) =\sqrt{\frac{(s_0 + T + \sqrt{a})^2 - a}{(s + T+\sqrt{a})^2 - a}}=\sqrt{\frac{(s_0+T)(s_0 + T + 2\sqrt{a})}{(s+T)(s + T+2\sqrt{a})}}$$

that is, $f^{(n)}(s_0)$, and I've get with Wolfram the following secuence

• $\displaystyle \frac{f'(s_0)}{1!}=\frac{ \left(\sqrt{a}+s_0+T\right)}{(s_0+T) \left(2 \sqrt{a}+s_0+T\right)}$
• $\displaystyle \frac{f''(s_0)}{2!}=\frac{ \left(\sqrt{a}+s_0+T\right)^2+\frac{a}{2}}{(s_0+T)^2 \left(2 \sqrt{a}+s_0+T\right)^2}$
• $\displaystyle \frac{f'''(s_0)}{3!}=\frac{\left(\left(\sqrt{a}+s_0+T\right)^2+\frac{3}{2}a \right)\left(\sqrt{a}+s_0+T\right)}{ (s_0+T)^3 \left(2 \sqrt{a}+s_0+T\right)^3}$

• $\displaystyle \frac{f''''(s_0)}{4!}=\frac{\left(\left(\sqrt{a}+s_0+T\right)^2+\frac{3}{2}a\right)^2-\frac{15}{8}a^2}{(s_0+T)^4 \left(2 \sqrt{a}+s_0+T\right)^4}$

• $\displaystyle \frac{f^{(5}(s_0)}{5!}=\frac{\left(\left(\left(\sqrt{a}+s_0+T\right)^2+\frac{5}{2}a\right)^2-\frac{35 a^2}{8}\right)\left(\sqrt{a}+s_0+T\right)}{(s_0+T)^5 \left(2 \sqrt{a}+s_0+T\right)^5}$

  • $\displaystyle \frac{f^{(6}(s_0)}{6!}=\frac{\left(\left(\left(\sqrt{a}+s_0+T\right)^2+\frac{15}{4}a\right)^2-\frac{135 }{16}a^2\right)\left(\sqrt{a}+s_0+T\right)^2+\frac{5 a^3}{16}}{(s_0+T)^6 \left(2 \sqrt{a}+s_0+T\right)^6}$

Can anyone see any relationship for $\displaystyle \frac{f^{(n}(s_0)}{n!}$?

It's clear that $\displaystyle \frac{f^{(n}(s_0)}{n!}=\frac{???????}{(s_0+T)^n \left(2 \sqrt{a}+s_0+T\right)^n}$ but the numerator is unknown for me.

Answer 1

You make life complex considering the numerator which is a constant. So $$f(s) =K\big({(s + T+\sqrt{a})^2 - a}\big)^{-\frac 12} \qquad \text{where} \qquad K=\sqrt{(s_0 + T + \sqrt{a})^2 - a}$$ To make it easier, let $A= T+\sqrt{a}$ making that we only need to consider $$g(s)=\frac{1}{\left((s+A)^2-a\right)^{1/2}}$$ and now, trying Taylor expansion around $s_0$, it seems that, at $s=s_0$, $\displaystyle \frac{g^{(n)}}{n!}$ is given by a recurrence relation $$(n+2) \left((s_0+A)^2-a\right)\,g^{(n+2)}+(2 n+3) (s_0+A)\,g^{(n+1)}+(n+1)\, g^{(n)}=0$$ with $$g^{(0)}=\frac{1}{\left((s_0+A)^2-a\right)^{1/2}}\qquad \text{and } \qquad g^{(1)}=-\frac{s_0+A}{\left((s_0+A)^2-a\right)^{3/2}}$$

Answer 2

This is not a full answer, merely suggestions on how to tackle this. As @Claude Leibovici has pointed out, you are carrying around too much baggage. Here are the simplifications I think you should make.

(0) Trivially, write $a=b^2$.

(1) Perform a translation to place the origin at $T+b$. This will make no difference to the higher derivatives. We will then be looking at the function $$ g(t) =\sqrt{\frac{t_0^2 - b^2}{t^2 - b^2}}. $$

(2) Scale, changing the variable to $bu$; it is easy to see what this does to the higher derivatives. We will then be looking at the function $$ h(u) =\sqrt{\frac{u_0^2 - 1}{u^2 - 1}}. $$

(3) Scale, replacing the function by a constant multiple; again it is clear what this does to derivatives. We'll then be looking at $$ k(u) =\frac{1}{\sqrt{u^2 - 1}}. $$

At this stage you have choices, and the real work begins. My own preference would be to change the variable by $u=\sec\theta$, our function then being $\cot\theta$. That tells us, by the way, something about the "natural" parameters of the underlying problem.