samples/texts/1085779/page_5.md

$$f^{(n)} = \frac{-x(-1)^n n!}{2(a-x^2)^{n+1}} \quad \text{and} \quad g^{(n)} = \frac{(-1)^n (2n-1)!!}{2^n a^{\frac{2n+1}{2}}}$$

on substituting the above values in eq 4 we'll get

$$(fg{)}^i=\frac{-x}{2}\sum _{k=0}^i\left(\genfrac{}{}{0px}{}{i}{k}\right)\frac{(-1{)}^i(i-k)!(2k-1)!!}{(a-x^2{)}^{i-k+1}{2}^k{a}^{\frac{2k+1}{2}}}(fg)^i\; =\; \backslash frac\{-x\}\{2\}\; \backslash sum\_\{k=0\}^\{i\}\; \backslash binom\{i\}\{k\}\; \backslash frac\{(-1)^i\; (i-k)!\; (2k-1)!!\}\{(a-x^2)^\{i-k+1\}\; 2^k\; a^\{\backslash frac\{2k+1\}\{2\}\}\}$$

since $i \ge 1$ we have

$$(fg{)}^i=\frac{-x}{2}\sum _{k=0}^{i-1}\left(\genfrac{}{}{0px}{}{i-1}{k}\right)\frac{(-1{)}^{i-1}(i-k-1)!(2k-1)!!}{(a-x^2{)}^{i-k}{2}^k{a}^{\frac{2k+1}{2}}}(fg)^i\; =\; \backslash frac\{-x\}\{2\}\; \backslash sum\_\{k=0\}^\{i-1\}\; \backslash binom\{i-1\}\{k\}\; \backslash frac\{(-1)^\{i-1\}(i-k-1)!(2k-1)!!\}\{(a-x^2)^\{i-k\}\; 2^k\; a^\{\backslash frac\{2k+1\}\{2\}\}\}$$

on rearranging and writing the double factorial in terms of Pochammer symbol[8]. i.e $\frac{(2k-1)!!}{2^k} = \left(\frac{1}{2}\right)_k$

we'll get

$$(fg{)}^i=\frac{x(-1{)}^i(i-1)!}{2\sqrt{a}(a-x^2{)}^i}\sum _{k=0}^{i-1}{(1-\frac{x^2}{a})}^k{\left(\frac{1}{2}\right)}_k(fg)^i\; =\; \backslash frac\{x\; (-1)^i\; (i-1)!\}\{2\backslash sqrt\{a\}(a-x^2)^i\}\; \backslash sum\_\{k=0\}^\{i-1\}\; \backslash left(1-\backslash frac\{x^2\}\{a\}\backslash right)^k\; \backslash left(\backslash frac\{1\}\{2\}\backslash right)\_k$$

$$\therefore (fg)=\frac{\mathrm{\partial }}{\mathrm{\partial }a}{\tanh }^{-1}\left(\frac{x}{\sqrt{a}}\right)\backslash therefore\; (fg)\; =\; \backslash frac\{\backslash partial\}\{\backslash partial\; a\}\; \backslash tanh^\{-1\}\; \backslash left(\; \backslash frac\{x\}\{\backslash sqrt\{a\}\}\; \backslash right)$$

$$\therefore \frac{{\mathrm{\partial }}^i}{\mathrm{\partial }{a}^i}{\tanh }^{-1}\left(\frac{x}{\sqrt{a}}\right)=\frac{x(-1{)}^i(i-1)!}{2\sqrt{a}(a-x^2{)}^i}\sum _{k=0}^{i-1}{(1-\frac{x^2}{a})}^k{\left(\frac{1}{2}\right)}_k\backslash therefore\; \backslash quad\; \backslash frac\{\backslash partial^i\}\{\backslash partial\; a^i\}\; \backslash tanh^\{-1\}\; \backslash left(\; \backslash frac\{x\}\{\backslash sqrt\{a\}\}\; \backslash right)\; =\; \backslash frac\{x\; (-1)^i\; (i-1)!\}\{2\backslash sqrt\{a\}(a-x^2)^i\}\; \backslash sum\_\{k=0\}^\{i-1\}\; \backslash left(1\; -\; \backslash frac\{x^2\}\{a\}\backslash right)^k\; \backslash left(\backslash frac\{1\}\{2\}\backslash right)\_k$$

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Proof of Theorem 1

Proof.