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Jul
22
revised $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
added 18 characters in body
Jul
21
comment $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
Thanks a lot for your answer, Is $c(z)=1$ for $z∈C$? Do you have any suggestion to prove that?
Jul
21
comment $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
@PinkElephants : Now, I have understood what you mean in your answer. "we cannot ignore the terms in the sum with k<0" . Yes it is my wrong suggestion. Thanks a lot for your comments and answer
Jul
21
comment $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
If $x=2$ then $c'(0) \approx -0.596347$ or not ? Could you please calculate it if it is easy for your calculation tool. Thanks a lot .
Jul
21
revised $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
edited title
Jul
21
revised $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
added 200 characters in body
Jul
21
comment $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
@DonAntonio Yes It is my own idea. I have not seen that in other place. Now I edited my results after Norbert's feedbacks. Thanks to Norbert for his good feedbacks.
Jul
21
revised $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
added 475 characters in body
Jul
21
comment $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
@Norbert I edited my result. Thanks for your advice
Jul
21
revised $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
added 63 characters in body
Jul
21
comment $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
@Norbert You can see in my results. I derivated $\frac{d}{dx}(\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)})$ and I got $e^x$.
Jul
21
asked $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?
Jul
21
comment To express $f(x,z)=\sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{n!}$ as known functions
@ChristianBlatter :If so, Which initial conditions are required to get the function above? Thanks
Jul
20
revised To express $f(x,z)=\sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{n!}$ as known functions
added 35 characters in body
Jul
20
revised To express $f(x,z)=\sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{n!}$ as known functions
added 140 characters in body
Jul
20
asked To express $f(x,z)=\sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{n!}$ as known functions
Jul
19
comment General solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$
@Sasha Wonderful answer. You wrote very nice results. It was extra nice to see the relation between Airy functions and Hyper-geometric series. Thanks a lot.
Jul
19
accepted General solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$
Jul
18
revised General solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$
added 19 characters in body
Jul
18
revised General solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$
added 1356 characters in body