nth derivative of: $F(x)=1-\sqrt{1-x^2}$

Question

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?

Answer 1

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.

Answer 2

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 $F(x)=1-g(x^2)$

Answer 3

Hint

Calculate $F'(x),F''(x),F^{(3)}(x)$ and try to find a pattern. Prove it using induction.

Answer 4

Let $v=v(x)=1-x^2$. For $k\ge0$, making use of the Faa di Bruno formula and some properties of the partial Bell polynomials $B_{n,k}$, we have \begin{align*} \bigl(\sqrt{1-x^2}\,\bigr)^{(k)} &=\sum_{j=0}^{k}\frac{\operatorname{d}^j\sqrt{v}\,}{\operatorname{d} v^j}B_{k,j}(-2x,-2,0,\dotsc,0)\\ &=\sum_{j=0}^{k}\biggl\langle\frac12\biggr\rangle_jv^{1/2-j} (-1)^j2^jB_{k,j}(x,1,0,\dotsc,0)\\ &=\sum_{j=0}^{k}\biggl\langle\frac12\biggr\rangle_j \bigl(1-x^2\bigr)^{1/2-j} (-1)^j2^j \frac{1}{2^{k-j}}\frac{k!}{j!}\binom{j}{k-j}x^{2j-k}\\ &=-\sqrt{1-x^2}\,\frac{k!}{(2x)^k} \sum_{j=0}^{k}\frac{2^{j}(2j-3)!!}{j!}\binom{j}{k-j}\frac{x^{2j}}{(1-x^2)^{j}}. \end{align*} Consequently, we acquire \begin{equation} F^{(k)}(x)=\bigl(1-\sqrt{1-x^2}\,\bigr)^{(k)} =\sqrt{1-x^2}\,\frac{k!}{(2x)^k} \sum_{j=0}^{k}\frac{2^{j}(2j-3)!!}{j!}\binom{j}{k-j}\frac{x^{2j}}{(1-x^2)^{j}}, \quad k\ge0. \end{equation}

References

1. Siqintuya Jin, Bai-Ni Guo, and Feng Qi, Partial Bell polynomials, falling and rising factorials, Stirling numbers, and combinatorial identities, Computer Modeling in Engineering & Sciences 132 (2022), no. 3, 781--799; available online at https://doi.org/10.32604/cmes.2022.019941.
2. F. Qi and B.-N. Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterr. J. Math. 14 (2017), no. 3, Article 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
3. F. Qi, D.-W. Niu, D. Lim, and Y.-H. Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, J. Math. Anal. Appl. 491 (2020), no. 2, Article 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.