anonymous
  • anonymous
Interesting pattern I stumbled upon while playing with \(p\)-series.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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 \(n\ge2\). \[\large\sum_{i=1}^n a_i=0\]
anonymous
  • 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...
anonymous
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Anyway, a few more derivatives to convince the non-believers. \[\begin{align*}\frac{d^4f_k}{dx^4}&=\frac{f_{k-4}-3f_{k-3}+2f_{k-2}}{x^4}-3\frac{f_{k-3}-3f_{k-2}+2f_{k-1}}{x^4}\\[1ex] &=\frac{f_{k-4}-6f_{k-3}+11f_{k-2}-6f_{k-1}}{x^4} \end{align*}\] \[\begin{align*}\frac{d^5f_k}{dx^5}&=\frac{f_{k-5}-6f_{k-4}+11f_{k-3}-6f_{k-2}}{x^5}-4\frac{f_{k-4}-6f_{k-3}+11f_{k-2}-6f_{k-1}}{x^5}\\[1ex] &=\frac{f_{k-5}-10f_{k-4}+35f_{k-3}-50f_{k-2}+24f_{k-1}}{x^5} \end{align*}\] \[\begin{align*}\frac{d^6f_k}{dx^6}&=\frac{f_{k-6}-10f_{k-5}+35f_{k-4}-50f_{k-3}+24f_{k-2}}{x^6}\\[1ex] &\quad\quad-5\frac{f_{k-5}-10f_{k-4}+35f_{k-3}-50f_{k-2}+24f_{k-1}}{x^6}\\[1ex] &=\frac{f_{k-6}-15f_{k-5}+85f_{k-4}-225f_{k-3}+274f_{k-2}-120f_{k-1}}{x^6} \end{align*}\]
anonymous
  • anonymous
In each case thus far, the sum of the numerator's coefficients turns out to be zero.
anonymous
  • 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 \((-1)^{k+1} k!\).
anonymous
  • anonymous
Darn, there's already a name for this function: http://mathworld.wolfram.com/Polylogarithm.html
anonymous
  • anonymous
Here we are: http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html
anonymous
  • 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...
anonymous
  • anonymous
Whoops, should be \[\large \frac{d^n\text{Li}_n(x)}{dx^n}=\frac{\displaystyle\sum_{i=1}^n s(n,i) \text{Li}_{n-i}(x)}{x^n}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.