nth derivative of $\frac{e^x−1}{x}$ (both taylor series and finite sum)

Question

Consider the functiong $g(x) =\frac{e^x−1}{x}$. Find a general formula for $g^{(n)}(x)$and prove that this formula is correct.

If you want it as a finite sum,

Based on guess and check, I think this one would work:

$$\frac{d^n}{dx^n}\frac{e^x−1}{x}=\frac{e^xn!(-1)^n+n!(-1)^{n+1}}{x^{n+1}}+ \frac{e^x}{x^{n+1}}\sum_{j=1}^{n}\frac{x^jn!(-1)^{n-j}}{j!}$$

Answer 1

$e^x = \sum_\limits{n=0}^\infty \frac{x^n}{n!}\\ e^x - 1 = \sum_\limits{n=1}^\infty \frac{x^n}{n!}\\ \frac {e^x - 1}{x} = \sum_\limits{n=0}^\infty \frac{x^n}{(n+1)!}\\ \frac {d}{dx}\frac {e^x - 1}{x} = \sum_\limits{n=1}^\infty \frac{nx^{n-1}}{(n+1)!}=\sum_\limits{n=0}^\infty \frac{(n+1)x^{n}}{(n+2)!}=\sum_\limits{n=0}^\infty \frac{x^{n}}{(n+2)n!}\\ \frac {d^k}{dx^k}\frac {e^x - 1}{x} = \sum_\limits{n=0}^\infty \frac{x^{n}}{(n+k+1)n!}$

Answer 2

Hint:

From $$xy=e^x-1,$$ you draw

$$y+xy'=e^x,$$ $$2y'+xy''=e^x,$$ $$\cdots$$ $$ny^{(n-1)}+xy^{(n)}=e^x.$$

This gives you a way to check your formula.

At the same time, it hints you that the derivative will be proportional to $e^x$ and $x^{-n}$. A possible form is

$$\frac{P_n(x)}{x^{n+k}}e^x$$ when $P_n$ is a polynomial.

Answer 3

Continuing after answer - Doug M (https://math.stackexchange.com/users/317176/doug-m), nth derivative of $\frac{e^x−1}{x}$, URL (version: 2019-05-30): https://math.stackexchange.com/q/3245056

Let $$ \frac {d^k}{dx^k}\frac {e^x - 1}{x} = \sum_\limits{n=0}^\infty \frac{x^{n}}{(n+k+1)n!} = D_k $$ then, $$ D_k=\sum_\limits{n=0}^\infty \frac{x^{n}}{(n+k+1)n!} $$ $$ D_k=\frac{1}{(k+1)}\sum_\limits{n=0}^\infty \frac{(k+1)x^{n}}{(n+k+1)n!}=\frac{1}{(k+1)}\sum_\limits{n=0}^\infty \frac{(k+1+n-n)x^{n}}{(n+k+1)n!} $$ $$ D_k=\frac{1}{(k+1)}\sum_\limits{n=0}^\infty \left (\frac{k+1+n}{k+1+n}-\frac{n}{k+1+n} \right)\frac{x^{n}}{n!} $$ $$ D_k=\frac{1}{(k+1)}\sum_\limits{n=0}^\infty \frac{x^{n}}{n!}-\frac{1}{(k+1)}\sum_\limits{n=0}^\infty \frac{nx^{n}}{(n+k+1)n!} $$ $$ D_k=\frac{e^x}{(k+1)}-\frac{1}{(k+1)}\sum_\limits{n=1}^\infty \frac{nx^{n}}{(n+k+1)n!} $$ $$ D_k=\frac{e^x}{(k+1)}-\frac{1}{(k+1)}\sum_\limits{n=1}^\infty \frac{x^{n-1}x}{((n-1)+(k+1)+1)(n-1)!} $$ $$ D_k=\frac{e^x}{(k+1)}-\frac{x}{(k+1)}\sum_\limits{n=0}^\infty \frac{x^n}{(n+(k+1)+1)n!} $$ $$ D_k=\frac{e^x}{(k+1)}-\frac{x}{(k+1)}D_{k+1} $$ $$ xD_{k+1}+(k+1)D_k=e^x $$ or $$ xD_k+kD_{k-1}=e^x $$ Solve this to get a closed form of $D_k$.

Answer 4

Another way to put it $$ \eqalign{ & {{e^{\,x} - 1} \over x} = \sum\limits_{1\, \le \;n} {{{x^{\,n - 1} } \over {n!}}} = \sum\limits_{0\, \le \;n} {{{x^{\,n} } \over {\left( {n + 1} \right)!}}} = \sum\limits_{0\, \le \;n} {{1 \over {\left( {n + 1} \right)}}{{x^{\,n} } \over {n!}}} = {1 \over x}\int_{t = 0}^x {e^{\,t} dt} \cr & {{d^{\,m} } \over {dx^{\,m} }}{{e^{\,x} - 1} \over x} = \sum\limits_{m\, \le \;n} {{{n\left( {n - 1} \right) \cdots \left( {n - m + 1} \right)x^{\,n - m} } \over {\left( {n + 1} \right)!}}} = \cr & = \sum\limits_{m\, \le \;n} {{1 \over {n + 1}}{{x^{\,n - m} } \over {\left( {n - m} \right)!}}} = \sum\limits_{0\, \le \;n} {{1 \over {n + m + 1}}{{x^{\,n} } \over {n!}}} = \cr & = {1 \over {x^{\,m} }}\sum\limits_{0\, \le \;n} {{1 \over {n + m + 1}}x^{\,m} {{x^{\,n} } \over {n!}}} = {1 \over {x^{\,m} }}\int_{t = 0}^x {t^{\,m} e^{\,t} dt} \cr} $$

If you want you can relate the integral to the incomplete Gamma function ...