Prove that: xlog⁡(x)=(x−1x)+32!(x−1x)2+113!(x−1x)3+…Snn!(x−1x)n Where Sn= absolute Striling Numbers of first kind (0,1,3,11,50,274…)

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Prove that:  xlog⁡(x)=(x−1x)+32!(x−1x)2+113!(x−1x)3+…Snn!(x−1x)n  Where Sn= absolute Striling Numbers of first kind (0,1,3,11,50,274…)

tsjutten20 · Accepted Answer

log⁡(1−x)=−∑n=1∞1nxn  so  log⁡(x)=−log⁡1x=−log⁡(1−x−1x)=∑n=1∞1n(x−1x)n  And  11−x=∑n=0∞xn  so  x=11−x−1x=∑n=0∞(x−1x)n  and thus  xlog⁡(x)=(∑n=1∞1n(x−1x)n)⋅∑m=0∞(x−1x)m=∑n=1∞an(x−1x)n  where an=∑k=0n−11n−k=1+12+13+…1n=Snn!  So the result is  xlog⁡(x)=∑n=1∞Snn!(x−1x)n

ataill0k · Accepted Answer

we will need the fact that  n!Hn=n!(1+12+13+…+1n)=S(n,2)  where S(n,2) is the Stirlings number of the first kind  let me make a change of variable u=x−1x, x=11−u. Then  xln⁡x=11−uln⁡(11−u)=−11−uln⁡(1−u)  =(1+u+u2+…)(u+u22+u33+…)  =u+(12+1)u2+(13+12+1)u3+(14+13+12+1)u4+…  =u+H2u2+H3u3+H4u4  =u+S(2,2)2!u2+S(3,2)3!u3+S(4,2)4!u4+…+S(n,2)n!un+…

RizerMix · Accepted Answer

We have xlog⁡(x)=x[−log⁡(1/x)]=x[−log⁡(1−x−1x)]=x∑m=0∞1m+1(x−1x)m+1 and x=11x=11−x−1x=∑m=0∞(x−1x)m Thus xlog⁡(x)=∑m=0∞(x−1x)m∑m=0∞1m(x−1x)m=∑m=0∞∑k=0m1k+1(x−1x)m+1 which is the same as ∑m=0∞Smm!(x−1x)m+1

Prove that: x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_n}{n!}(\frac{x-1}{x})^n Where S_n= absolute Striling Numbers of first kind (0,1,3,11,50,274...)

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