anonymous one year ago Interesting pattern I stumbled upon while playing with $$p$$-series.

1. anonymous

I'll define a function, $\large f_k(x):=\sum_{n=1}^\infty \frac{x^n}{n^k}$ which converges for $$|x|<1$$. Consider its first derivative: \large\begin{align*}\frac{df_k}{dx}&=\sum_{n=1}^\infty \frac{nx^{n-1}}{n^k}\\[1ex] &=\frac{1}{x}\sum_{n=1}^\infty \frac{x^n}{n^{k-1}}\\[1ex] &=\frac{f_{k-1}}{x}\end{align*} For (seemingly) every derivative beyond this, a neat pattern emerges. Take the second derivative: \large\begin{align*}\frac{d^2f_k}{dx^2}&=\frac{d}{dx}\left[\frac{f_{k-1}}{x}\right]\\[1ex] &=\frac{f_{k-2}-f_{k-1}}{x^2}\end{align*} and the third derivative: \large\begin{align*}\frac{d^3f_k}{dx^3}&=\frac{f_{k-3}-3f_{k-2}+2f_{k-1}}{x^3}\end{align*} and so on. (I'm writing the results as I seem to remember them, they could be wrong but easy enough to work out) As you continue this process, you end up with this $$k$$th term formula: $\large\frac{d^nf_k}{dx^n}=\frac{\displaystyle\sum_{i=1}^na_if_{k-i}}{x^k}$ The "neat" part, at least in my opinion, is that this seems to hold for all $$n\ge2$$. $\large\sum_{i=1}^n a_i=0$

2. anonymous

I haven't checked for orders of the derivative larger than $$5$$, so it's possible that it fails beyond that. But if it doesn't, I wonder if it's a trivial consequence of the derivatives of the reciprocal powers, or if there's something more going on here...

3. anonymous

On another note, is there a class of differential equations that involve DEs as recurrence relations? Like a combo of difference and differential eqs.

4. anonymous

Anyway, a few more derivatives to convince the non-believers. \begin{align*}\frac{d^4f_k}{dx^4}&=\frac{f_{k-4}-3f_{k-3}+2f_{k-2}}{x^4}-3\frac{f_{k-3}-3f_{k-2}+2f_{k-1}}{x^4}\\[1ex] &=\frac{f_{k-4}-6f_{k-3}+11f_{k-2}-6f_{k-1}}{x^4} \end{align*} \begin{align*}\frac{d^5f_k}{dx^5}&=\frac{f_{k-5}-6f_{k-4}+11f_{k-3}-6f_{k-2}}{x^5}-4\frac{f_{k-4}-6f_{k-3}+11f_{k-2}-6f_{k-1}}{x^5}\\[1ex] &=\frac{f_{k-5}-10f_{k-4}+35f_{k-3}-50f_{k-2}+24f_{k-1}}{x^5} \end{align*} \begin{align*}\frac{d^6f_k}{dx^6}&=\frac{f_{k-6}-10f_{k-5}+35f_{k-4}-50f_{k-3}+24f_{k-2}}{x^6}\\[1ex] &\quad\quad-5\frac{f_{k-5}-10f_{k-4}+35f_{k-3}-50f_{k-2}+24f_{k-1}}{x^6}\\[1ex] &=\frac{f_{k-6}-15f_{k-5}+85f_{k-4}-225f_{k-3}+274f_{k-2}-120f_{k-1}}{x^6} \end{align*}

5. anonymous

In each case thus far, the sum of the numerator's coefficients turns out to be zero.

6. anonymous

For another thing, if we wanted to compute the $$k$$th-order derivative, there's at least an obvious pattern with the coefficients of the first and last terms, namely $$1$$ and $$(-1)^{k+1} k!$$.

7. anonymous

Darn, there's already a name for this function: http://mathworld.wolfram.com/Polylogarithm.html

8. anonymous
9. anonymous

So if $f_k(x)\equiv \text{Li}_k(x)=\sum_{n=1}^\infty \frac{x^n}{n^k}$ then it looks like we have $\large \frac{d^n\text{Li}_n(x)}{dx^n}=\frac{\displaystyle\sum_{i=1}^n s(n,i) \text{Li}_i(x)}{x^n}$ And apparently (and Mathematica concurs) $\sum_{i=1}^ns(n,i)=0$ It's interesting that this last identity isn't listed on MathWorld...

10. anonymous

Whoops, should be $\large \frac{d^n\text{Li}_n(x)}{dx^n}=\frac{\displaystyle\sum_{i=1}^n s(n,i) \text{Li}_{n-i}(x)}{x^n}$