0, ~~~\forall i\in N^{+}$

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when toggle format	what		by	license	comment
S Apr 14, 2021 at 0:06	history	bounty ended	math110
S Apr 14, 2021 at 0:06	history	notice removed	math110
Apr 14, 2021 at 0:05	vote	accept	math110
Apr 9, 2021 at 7:44	answer	added	Mateusz Kwaśnicki		timeline score: 2
Apr 9, 2021 at 4:28	history	edited	math110	CC BY-SA 4.0	deleted 1 character in body
S Apr 9, 2021 at 4:18	history	bounty started	math110
S Apr 9, 2021 at 4:18	history	notice added	math110		Authoritative reference needed
Apr 9, 2021 at 4:17	comment	added	math110		@IraGessel,Now How to prove coefficients of $1-\left(-\dfrac{x}{\ln{(1-x)}}\right)^{\frac{1}{K}}$ are postive ?can you solve it? Thanks
Apr 9, 2021 at 4:16	comment	added	math110		@IraGessel,oh, I known,. Thank you,
Apr 9, 2021 at 4:16	history	edited	math110	CC BY-SA 4.0	added 4 characters in body
Apr 9, 2021 at 4:15	comment	added	Ira Gessel		$a_n$ is positive only for $n>0$; $a_0=-1$. Here is a suggestion: in general, differentiating and then setting $x=0$ is not an efficient way of computing Taylor series coefficients. It is usually more work to compute the derivatives than to compute the Taylor series in some other way. For example, it's very easy to compute the Taylor series for $e^{x^2}$ by substituting $x^2$ for $x$ in $e^x = 1+x + x^2/2!+\cdots$, but it's more complicated to compute the derivatives of $e^{x^2}$.
Apr 9, 2021 at 2:55	history	edited	math110	CC BY-SA 4.0	added 118 characters in body
Apr 9, 2021 at 2:50	history	edited	math110	CC BY-SA 4.0	added 118 characters in body
Apr 9, 2021 at 2:47	comment	added	math110		why is $1-\left(-\dfrac{x}{\ln{(1-x)}}\right)^{\frac{1}{K}}$, not $\left(-\dfrac{x}{\ln{(1-x)}}\right)^{\frac{1}{K}}$?Where did that $1?$ come from, Thanks
Apr 7, 2021 at 18:10	comment	added	Ira Gessel		This question was also asked at math.stackexchange.com/questions/3983857/….
Apr 7, 2021 at 18:03	comment	added	Ira Gessel		The problem is equivalent to showing that the Taylor series coefficients of $$1-\left(-\frac{x}{\ln(1-x)}\right)^{1/K} =\frac{x}{2K} +\frac{5K-3}{24K^2}x^2+\cdots $$ are positive when $K$ is a positive integer.
Apr 7, 2021 at 17:13	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 12 characters in body
Apr 7, 2021 at 17:04	answer	added	Iosif Pinelis		timeline score: 2
Apr 7, 2021 at 15:08	comment	added	math110		@IosifPinelis have edit it.can you see where I am wrong
Apr 7, 2021 at 15:06	history	edited	math110	CC BY-SA 4.0	added 809 characters in body
Apr 7, 2021 at 5:58	history	edited	gmvh	CC BY-SA 4.0	Added top-level tag
Apr 7, 2021 at 0:38	comment	added	Gerry Myerson		Indeed, for $0<x<1$, we have $x/\log(1-x)<0$, so $(x/\log(1-x))^{1/K}$ isn't real for, say, $K=2$.
Apr 7, 2021 at 0:32	comment	added	Iosif Pinelis		For $f(x):=\left(\dfrac{x}{\ln (1-x)}\right)^{1/K}$, we have $f'(0+)=-\frac{1}{2K} (-1)^{1/K}$. How do you want it to be $>0$? Also, what is $N^+$?
Apr 7, 2021 at 0:01	history	edited	math110	CC BY-SA 4.0	added 35 characters in body
Apr 6, 2021 at 23:55	history	asked	math110	CC BY-SA 4.0

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