Sum with a fractional number of terms???

Question

I was playing around on desmos with the Bernoulli polynomials which are defined as $$B(x,m)=\sum_{n=0}^{m}\frac1{n+1}\sum_{k=0}^{n}(-1)^k{n\choose k}(x+k)^m$$ And I noticed that graphs were being displayed for fractional $m$. I guess that is to be expected, as non-integer $m$ is usually interpreted as $\lfloor m\rfloor$. But the confusing part, as you will see in the link I provided, is that desmos thinks that $$B(x,5/3)\neq B(x,7/4)$$ even though $\lfloor5/3\rfloor=\lfloor7/4\rfloor=1$. I am confused. Could someone explain how summation can be extended to non-integer indices?

Answer 1

The interpretation that you give is in fact the correct one for a sum indicated as, e.g. $$ \sum\limits_{0\, \le \,n\, \le \,3/2} {f(n)} = \sum\limits_{0\, \le \,n\, \le \,\left\lfloor {3/2} \right\rfloor } {f(n)} = f(0) + f(1) $$

But there is a definition of Sum that applies to fractional, real, or even complex indices. That is based on the concept of Antidelta or Indefinite Summation.

If we can establish that $$ f(n) = F(n + 1) - F(n) = \Delta F(n) $$ then we have that $$ \eqalign{ & F(n) = \Delta ^{\left( { - 1} \right)} f(n) = \sum\nolimits_{k = 0}^n {f(k)} + c = \sum\limits_{0\, \le \,k\, \le \,n - 1} {f(k)} + c\quad \Rightarrow \cr & \Rightarrow \sum\limits_{m\, \le \,k\, \le \,n - 1} {f(k)} = \sum\nolimits_{k = m}^n {f(k)} = F(n) - F(m) \cr} $$

So, if $f(z), F(z)$ exist also for real or complex $z$, then it is natural to define $$ \sum\nolimits_{\,k = a\;}^{\,b} {f(k)} = F(b) - F(a) $$

For example $$ \eqalign{ & F(x) = {{x\left( {x - 1} \right)} \over 2}\quad \Rightarrow \quad \Delta F(x) = {{\left( {x + 1} \right)x} \over 2} - {{x\left( {x - 1} \right)} \over 2} = x \cr & \sum\nolimits_{k = 0}^x k = {{x\left( {x - 1} \right)} \over 2} = \left( \matrix{ x \cr 2 \cr} \right) = {{\Gamma (x + 1)} \over {\Gamma (3)\Gamma (x - 1)}} \cr} $$

Regarding the Bernoulli polynomials the indicated link provides various relations that can be used to extend their definition to non-integral indices, including $$ \sum\nolimits_{k = 0}^x {k^s } = {{B_{\,s + 1} (x) - B_{\,s + 1} (0)} \over {s + 1}} $$