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.

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