I've gotten this function from probability generating functions, and I want to calculate it's nth derivative (With respect to $x$).
This is: $$F(x)=1-\sqrt{1-x^2}$$
Is there a practical way to do it?
Or for another approach, I just need the derivatives calculated in $x=0$, to calculate it's MacLaurin series. Any practical way to do it?
$\endgroup$ 13 Answers
$\begingroup$Expand $\sqrt{1-x^2}$ using the binomial theorem: you'll get something like $$ \sqrt{1-x^2}=\sum_{k\geqslant 0} (-x)^{2k} \binom{1/2}{k}, $$ which you can then expand out into a product.
$\endgroup$ 1 $\begingroup$let $g(x)=(1-x)^\frac 12$
$$g'(x)=-\frac 12 (1-x)^{-\frac 12} \\ g''(x)=-\frac 12 \frac 12 (1-x)^{-\frac 32} \\ g'''(x)=-\frac 32 \frac 12 \frac 12 (1-x)^{-\frac 52} \\ g^{(4)}(x)=-\frac 52\frac 32 \frac 12 \frac 12 (1-x)^{-\frac 72} \\ g^{(5)}(x)=-\frac 72\frac 52\frac 32 \frac 12 \frac 12 (1-x)^{-\frac 92} \\ $$
so $$g^{(n)}(x)=-\frac {1}{2^n} (2n-3)(2n-5)(2n-7) ... (5)(3)(1) (1-x)^{-\frac{2n-1}{2} } \\ $$
$$g^{(n)}(0)=-\frac {1}{2^n} (2n-3)(2n-5)(2n-7) ... (5)(3)(1) \\ = -\frac {1}{2^n} \frac{(2n-3)!}{(2n-2)(2n-4)(2n-6) ... (4)(2)} \\ = -\frac {1}{2^n} \frac{(2n-3)!}{2^{n-1}(n-1)!} \\ = -\frac {1}{2^{2n-1}} \frac{(2n-3)!}{(n-1)!} $$
You should be able to complete this knowing that $F(x)=1-g(x^2)$
$\endgroup$ 1 $\begingroup$Hint
Calculate $F'(x),F''(x),F^{(3)}(x)$ and try to find a pattern. Prove it using induction.
$\endgroup$
