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qifeng618
qifeng618
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In the references below, the derivative formula \begin{equation}\label{exp-frac1x-expans}\tag{QF1} \bigl(\operatorname{e}^{\pm1/t}\bigr)^{(n)} =(-1)^n{\operatorname{e}^{\pm1/t}}\sum_{k=0}^{n}(\pm1)^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0 \end{equation} was derived alternatively. From \eqref{exp-frac1x-expans}, it follows that \begin{equation}\label{exp-frac1x-alpha}\tag{QF2} \bigl(\operatorname{e}^{\alpha/t}\bigr)^{(n)} =(-1)^n\operatorname{e}^{\alpha/t}\sum_{k=0}^{n}\alpha^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0. \end{equation} Taking $\alpha=-\frac{\lambda}{2}$ in \eqref{exp-frac1x-alpha} gives \begin{equation*} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr)\biggr]^{(k)} =(-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{t^{k+\ell}}, \quad k\ge0. \end{equation*} Therefore, in light of the Leibnitz rule for differentiation, we obtain \begin{gather*} \begin{aligned} f^{(n)}(x)&=\biggl[\exp\biggl(\frac{\lambda}{\mu}\biggr) \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(n)}\\ &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k}\biggl[\exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr)\biggr]^{(n-k)} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(k)} \end{aligned}\\ \begin{aligned} &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k} \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=\exp\biggl(\frac{\lambda}{\mu}-\frac{\lambda}{2\mu^2}x-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{k=0}^{n}\binom{n}{k} \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0. \end{aligned} \end{gather*} HoweverIn conclusion, numerical computation by the software Mathematica shows thatwe obtain the derivative formula \begin{equation*} \boxed{f^{(n)}(x)=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0} \end{equation*} is not correct. But I cannot find any error in the above proof. What's wrong? How's wrong?\begin{equation*} \boxed{f^{(n)}(x)=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0.} \end{equation*}

References

  1. S. Daboul, J. Mangaldan, M. Z. Spivey, and P. J. Taylor, The Lah numbers and the $n$th derivative of $e^{1/x}$, Math. Mag. 86 (2013), no. 1, 39--47; Available online at http://dx.doi.org/10.4169/math.mag.86.1.039.
  2. F. Qi, An explicit formula for the Bell numbers in terms of the Lah and Stirling numbers, Mediterr. J. Math. 13 (2016), no. 5, 2795--2800; available online at https://doi.org/10.1007/s00009-015-0655-7.
  3. X.-J. Zhang, F. Qi, and W.-H. Li, Properties of three functions relating to the exponential function and the existence of partitions of unity, Int. J. Open Probl. Comput. Sci. Math. 5 (2012), no. 3, 122--127; available online at https://doi.org/10.12816/0006128.
  4. https://qifeng618.wordpress.com/2018/05/10/some-papers-related-to-the-function-e1-x-and-the-lah-numbers/

In the references below, the derivative formula \begin{equation}\label{exp-frac1x-expans}\tag{QF1} \bigl(\operatorname{e}^{\pm1/t}\bigr)^{(n)} =(-1)^n{\operatorname{e}^{\pm1/t}}\sum_{k=0}^{n}(\pm1)^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0 \end{equation} was derived alternatively. From \eqref{exp-frac1x-expans}, it follows that \begin{equation}\label{exp-frac1x-alpha}\tag{QF2} \bigl(\operatorname{e}^{\alpha/t}\bigr)^{(n)} =(-1)^n\operatorname{e}^{\alpha/t}\sum_{k=0}^{n}\alpha^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0. \end{equation} Taking $\alpha=-\frac{\lambda}{2}$ in \eqref{exp-frac1x-alpha} gives \begin{equation*} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr)\biggr]^{(k)} =(-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{t^{k+\ell}}, \quad k\ge0. \end{equation*} Therefore, in light of the Leibnitz rule for differentiation, we obtain \begin{gather*} \begin{aligned} f^{(n)}(x)&=\biggl[\exp\biggl(\frac{\lambda}{\mu}\biggr) \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(n)}\\ &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k}\biggl[\exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr)\biggr]^{(n-k)} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(k)} \end{aligned}\\ \begin{aligned} &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k} \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=\exp\biggl(\frac{\lambda}{\mu}-\frac{\lambda}{2\mu^2}x-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{k=0}^{n}\binom{n}{k} \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0. \end{aligned} \end{gather*} However, numerical computation by the software Mathematica shows that the derivative formula \begin{equation*} \boxed{f^{(n)}(x)=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0} \end{equation*} is not correct. But I cannot find any error in the above proof. What's wrong? How's wrong?

References

  1. S. Daboul, J. Mangaldan, M. Z. Spivey, and P. J. Taylor, The Lah numbers and the $n$th derivative of $e^{1/x}$, Math. Mag. 86 (2013), no. 1, 39--47; Available online at http://dx.doi.org/10.4169/math.mag.86.1.039.
  2. F. Qi, An explicit formula for the Bell numbers in terms of the Lah and Stirling numbers, Mediterr. J. Math. 13 (2016), no. 5, 2795--2800; available online at https://doi.org/10.1007/s00009-015-0655-7.
  3. X.-J. Zhang, F. Qi, and W.-H. Li, Properties of three functions relating to the exponential function and the existence of partitions of unity, Int. J. Open Probl. Comput. Sci. Math. 5 (2012), no. 3, 122--127; available online at https://doi.org/10.12816/0006128.
  4. https://qifeng618.wordpress.com/2018/05/10/some-papers-related-to-the-function-e1-x-and-the-lah-numbers/

In the references below, the derivative formula \begin{equation}\label{exp-frac1x-expans}\tag{QF1} \bigl(\operatorname{e}^{\pm1/t}\bigr)^{(n)} =(-1)^n{\operatorname{e}^{\pm1/t}}\sum_{k=0}^{n}(\pm1)^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0 \end{equation} was derived alternatively. From \eqref{exp-frac1x-expans}, it follows that \begin{equation}\label{exp-frac1x-alpha}\tag{QF2} \bigl(\operatorname{e}^{\alpha/t}\bigr)^{(n)} =(-1)^n\operatorname{e}^{\alpha/t}\sum_{k=0}^{n}\alpha^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0. \end{equation} Taking $\alpha=-\frac{\lambda}{2}$ in \eqref{exp-frac1x-alpha} gives \begin{equation*} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr)\biggr]^{(k)} =(-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{t^{k+\ell}}, \quad k\ge0. \end{equation*} Therefore, in light of the Leibnitz rule for differentiation, we obtain \begin{gather*} \begin{aligned} f^{(n)}(x)&=\biggl[\exp\biggl(\frac{\lambda}{\mu}\biggr) \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(n)}\\ &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k}\biggl[\exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr)\biggr]^{(n-k)} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(k)} \end{aligned}\\ \begin{aligned} &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k} \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=\exp\biggl(\frac{\lambda}{\mu}-\frac{\lambda}{2\mu^2}x-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{k=0}^{n}\binom{n}{k} \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0. \end{aligned} \end{gather*} In conclusion, we obtain the derivative formula \begin{equation*} \boxed{f^{(n)}(x)=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0.} \end{equation*}

References

  1. S. Daboul, J. Mangaldan, M. Z. Spivey, and P. J. Taylor, The Lah numbers and the $n$th derivative of $e^{1/x}$, Math. Mag. 86 (2013), no. 1, 39--47; Available online at http://dx.doi.org/10.4169/math.mag.86.1.039.
  2. F. Qi, An explicit formula for the Bell numbers in terms of the Lah and Stirling numbers, Mediterr. J. Math. 13 (2016), no. 5, 2795--2800; available online at https://doi.org/10.1007/s00009-015-0655-7.
  3. X.-J. Zhang, F. Qi, and W.-H. Li, Properties of three functions relating to the exponential function and the existence of partitions of unity, Int. J. Open Probl. Comput. Sci. Math. 5 (2012), no. 3, 122--127; available online at https://doi.org/10.12816/0006128.
  4. https://qifeng618.wordpress.com/2018/05/10/some-papers-related-to-the-function-e1-x-and-the-lah-numbers/
Source Link
qifeng618
qifeng618
  • 1.1k
  • 5
  • 14

In the references below, the derivative formula \begin{equation}\label{exp-frac1x-expans}\tag{QF1} \bigl(\operatorname{e}^{\pm1/t}\bigr)^{(n)} =(-1)^n{\operatorname{e}^{\pm1/t}}\sum_{k=0}^{n}(\pm1)^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0 \end{equation} was derived alternatively. From \eqref{exp-frac1x-expans}, it follows that \begin{equation}\label{exp-frac1x-alpha}\tag{QF2} \bigl(\operatorname{e}^{\alpha/t}\bigr)^{(n)} =(-1)^n\operatorname{e}^{\alpha/t}\sum_{k=0}^{n}\alpha^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0. \end{equation} Taking $\alpha=-\frac{\lambda}{2}$ in \eqref{exp-frac1x-alpha} gives \begin{equation*} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr)\biggr]^{(k)} =(-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{t^{k+\ell}}, \quad k\ge0. \end{equation*} Therefore, in light of the Leibnitz rule for differentiation, we obtain \begin{gather*} \begin{aligned} f^{(n)}(x)&=\biggl[\exp\biggl(\frac{\lambda}{\mu}\biggr) \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(n)}\\ &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k}\biggl[\exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr)\biggr]^{(n-k)} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(k)} \end{aligned}\\ \begin{aligned} &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k} \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=\exp\biggl(\frac{\lambda}{\mu}-\frac{\lambda}{2\mu^2}x-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{k=0}^{n}\binom{n}{k} \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0. \end{aligned} \end{gather*} However, numerical computation by the software Mathematica shows that the derivative formula \begin{equation*} \boxed{f^{(n)}(x)=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0} \end{equation*} is not correct. But I cannot find any error in the above proof. What's wrong? How's wrong?

References

  1. S. Daboul, J. Mangaldan, M. Z. Spivey, and P. J. Taylor, The Lah numbers and the $n$th derivative of $e^{1/x}$, Math. Mag. 86 (2013), no. 1, 39--47; Available online at http://dx.doi.org/10.4169/math.mag.86.1.039.
  2. F. Qi, An explicit formula for the Bell numbers in terms of the Lah and Stirling numbers, Mediterr. J. Math. 13 (2016), no. 5, 2795--2800; available online at https://doi.org/10.1007/s00009-015-0655-7.
  3. X.-J. Zhang, F. Qi, and W.-H. Li, Properties of three functions relating to the exponential function and the existence of partitions of unity, Int. J. Open Probl. Comput. Sci. Math. 5 (2012), no. 3, 122--127; available online at https://doi.org/10.12816/0006128.
  4. https://qifeng618.wordpress.com/2018/05/10/some-papers-related-to-the-function-e1-x-and-the-lah-numbers/