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anonymous
 one year ago
Interesting pattern I stumbled upon while playing with \(p\)series.
anonymous
 one year ago
Interesting pattern I stumbled upon while playing with \(p\)series.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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: \[\large\begin{align*}\frac{df_k}{dx}&=\sum_{n=1}^\infty \frac{nx^{n1}}{n^k}\\[1ex] &=\frac{1}{x}\sum_{n=1}^\infty \frac{x^n}{n^{k1}}\\[1ex] &=\frac{f_{k1}}{x}\end{align*}\] For (seemingly) every derivative beyond this, a neat pattern emerges. Take the second derivative: \[\large\begin{align*}\frac{d^2f_k}{dx^2}&=\frac{d}{dx}\left[\frac{f_{k1}}{x}\right]\\[1ex] &=\frac{f_{k2}f_{k1}}{x^2}\end{align*}\] and the third derivative: \[\large\begin{align*}\frac{d^3f_k}{dx^3}&=\frac{f_{k3}3f_{k2}+2f_{k1}}{x^3}\end{align*}\] 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: \[\large\frac{d^nf_k}{dx^n}=\frac{\displaystyle\sum_{i=1}^na_if_{ki}}{x^k}\] The "neat" part, at least in my opinion, is that this seems to hold for all \(n\ge2\). \[\large\sum_{i=1}^n a_i=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0On another note, is there a class of differential equations that involve DEs as recurrence relations? Like a combo of difference and differential eqs.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anyway, a few more derivatives to convince the nonbelievers. \[\begin{align*}\frac{d^4f_k}{dx^4}&=\frac{f_{k4}3f_{k3}+2f_{k2}}{x^4}3\frac{f_{k3}3f_{k2}+2f_{k1}}{x^4}\\[1ex] &=\frac{f_{k4}6f_{k3}+11f_{k2}6f_{k1}}{x^4} \end{align*}\] \[\begin{align*}\frac{d^5f_k}{dx^5}&=\frac{f_{k5}6f_{k4}+11f_{k3}6f_{k2}}{x^5}4\frac{f_{k4}6f_{k3}+11f_{k2}6f_{k1}}{x^5}\\[1ex] &=\frac{f_{k5}10f_{k4}+35f_{k3}50f_{k2}+24f_{k1}}{x^5} \end{align*}\] \[\begin{align*}\frac{d^6f_k}{dx^6}&=\frac{f_{k6}10f_{k5}+35f_{k4}50f_{k3}+24f_{k2}}{x^6}\\[1ex] &\quad\quad5\frac{f_{k5}10f_{k4}+35f_{k3}50f_{k2}+24f_{k1}}{x^6}\\[1ex] &=\frac{f_{k6}15f_{k5}+85f_{k4}225f_{k3}+274f_{k2}120f_{k1}}{x^6} \end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In each case thus far, the sum of the numerator's coefficients turns out to be zero.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For another thing, if we wanted to compute the \(k\)thorder derivative, there's at least an obvious pattern with the coefficients of the first and last terms, namely \(1\) and \((1)^{k+1} k!\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Darn, there's already a name for this function: http://mathworld.wolfram.com/Polylogarithm.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here we are: http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whoops, should be \[\large \frac{d^n\text{Li}_n(x)}{dx^n}=\frac{\displaystyle\sum_{i=1}^n s(n,i) \text{Li}_{ni}(x)}{x^n}\]
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