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anonymous

  • one year ago

Interesting pattern I stumbled upon while playing with \(p\)-series.

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  1. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    In each case thus far, the sum of the numerator's coefficients turns out to be zero.

  6. anonymous
    • one year ago
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    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
    • one year ago
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    Darn, there's already a name for this function: http://mathworld.wolfram.com/Polylogarithm.html

  8. anonymous
    • one year ago
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    Here we are: http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html

  9. anonymous
    • one year ago
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    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
    • one year ago
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    Whoops, should be \[\large \frac{d^n\text{Li}_n(x)}{dx^n}=\frac{\displaystyle\sum_{i=1}^n s(n,i) \text{Li}_{n-i}(x)}{x^n}\]

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