Saturday, 27 August 2016

calculus - Power series representation




I'm trying to find the series representation of f(x)=x0et1+tdt. I have found it using the Maclaurin series, differentiating multiple times and finding a pattern. But I think there must be an easier way, using the power series of elementary functions. I know that ex=0xnn! and 11+x=(1)nxn but I don't know how to use it here. Thanks



(Don't hesitate to correct my English)


Answer



ex=n0xnn!



1x+1=n0(1)nxn



exx+1=b0a0(1)ba!xa+b




\frac{e^x}{x+1}=\sum_{n\ge 0}\sum_{a+b=n}_{a,b\ge 0} \frac{(-1)^b}{a!}x^{a+b}



exx+1=n=0xn(1)n(nk=0(1)k1k!)






Now the interesting thing about these coefficients above that are given in the form of partial sums is that it turns out that we have:



nk=0(1)kk!=1e+1n!((1)n+12{n!e})




Where {.} is the fractional part function.



So that we get:



exx+1=n=0xn(1)n(1e+1n!((1)n+12{n!e}))



=1en=0xn(1)n+n=0xn(1)n1n!((1)n+12{n!e})




=1e(x+1)+12n=0xnn!(1+(1)n)n=0xn(1)nn!{n!e}
=1e(x+1)+ex+ex2n=0xn(1)nn!{n!e}
=1e(x+1)+ex+ex2+n=0xnn!{n!e}(1)n+1



Now integrating term by term gives,



x0et1+tdt=ln(x+1)e+exex2+n=0xn+1(n+1)!{n!e}(1)n+1







Thus for all |x|<1 the following expansion holds:



f(x)=x0et1+tdt=ln(x+1)e+exex2+n=1(x)nn!{(n1)!e}



Also pretty off topic, but the positive nth coefficients that took the form of partial sums in our first list of series manipulations above correspond to solutions to the hat-check problem when n hats are being considered.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...