Using Bernoulli functions

Bernoulli polynomials (MathWorld, Wikipedia) are an extension of Bernoulli numbers. They obey this recurrence relation: $B_{n}(x)=(B+x)^{n}$

where $B^{n}$ is replaced by $B_{n}$ after the right-hand side has been expanded symbolically.

They can also be derived as coefficients in the following power series expansion: $\frac{te^{tx}}{e^{t}-1}=\sum_{n=0}^{\infty}B_{n}(x)\frac{t^{n}}{n!}$

The hopeful fact about Bernoulli polynomials in our case is that $\sum_{i=a}^{b}i^{n}=\frac{1}{n+1}\{ B_{n+1}(b+1)-B_{n+1}(a)\}$,

which means that the equation whose solutions we are investigating, $\sum_{1}^{n}(x-i)^{n}=x^{n}$

boils down to $\frac{1}{n+1}\{ B_{n+1}(x)-B_{n+1}(x-n)\}=x^{n}$

or even $B_{n+1}(x)-B_{n+1}(x-n)=(n+1)x^{n}$.

Now $B_{n+1}(x)$ is a polynomial of degree n+1 in x, so that in the area we’re looking in, where x-n is approximately 0.3x, $B_{n+1}(x-n)$ will be infinitesimal in comparison with $B_{n+1}(x)$, so that the simplified equation $B_{n+1}(x)=(n+1)x^{n}$

has the same asymptotic behaviour as the original.

Another hopeful line of inquiry is that $B_{n+1}(x)$ is the coefficient of $\frac{t^{n+1}}{(n+1)!}$in the power series expansion of $\frac{te^{tx}}{(e^{t}-1)}$, while $(n+1)x^{n}$ is the coefficient of $\frac{t^{n+1}}{(n+1)!}$ in the expansion of $te^{tx}$.

Unfortunately the the x we are looking for is not constant but depends on n, so that we can’t establish interesting facts about $e^{tx}/(e^{t}-1)$ and $e^{tx}$ as a whole. All we can really do to isolate a particular power of t is to differentiate each of these two functions n times with respect to t and compare the results at t=0. Nevertheless, it is still good to find some sort of an occurrence of an exponential function, given that we are trying to get a reason for ln 2 appearing in the result.