Lets say I have a sum that states the following $$ \sum_{j=0}^{k-c} {k-c \choose j}\ln(a)^{k-c-j} \frac{d^j}{dx^j}[(x)_c] $$ where $(x)_c$ is the falling factorial such that

$$ (x)_c = x(x-1)(x-2)\cdots(x-c+1) $$

How can I evaluate this summation?

Viewing the summations partially, I know that

$$ \sum_{j=0}^{k-c}{k-c \choose j}\ln(a)^{k-c-j} = \left(\frac{1}{\ln(a)}+1\right)^{k-c} \ln(a)^{k-c} $$ But it is possible that $k-c > c$ and therefore having the summation above be incomplete due to differentiating a constant, resulting in an incomplete gamma function result seen here: http://www.wolframalpha.com/input/?i=sum+%28%28k-c%29%21%2F%28%28k-c-r%29%21%29%29ln%28y%29%5E%28k-c-r%29%2C+r%3D0..k-c%2Bn

An interesting thing to note is that taking multiple derivatives of the falling factorial x results in the sum of the permutations that the falling factorial can be presented in, multiplied by the factorial of the amount of times differentiated. an example would be: $$ \frac{d}{dx} [x(x-1)(x-2)(x-3)] = 1![x(x-1)(x-2)+x(x-1)(x-3)+x(x-2)(x-3)+(x-1)(x-2)(x-3)] $$

$$ \frac{d^2}{dx^2} [x(x-1)(x-2)(x-3)] = 2![x(x-1)+x(x-2)+x(x-3)+(x-1)(x-2)+(x-1)(x-3)+(x-2)(x-3)] $$ Note that the number of terms produced by taking $n$ derivatives of $(x)_c$ is directly proportional to ${n \choose c}$

Is there no defined way to express this summation?

Thank you for any information in advance.

  • $\begingroup$ This puzzles me: On Wikipedia in numerous articles I've seen things like "$a+b+c$" Where $a$ is the number of things" etc., with a capital initial "W" in where, and someone told me that capitalization of the "W" was done by MS Word software, which mistakes the transition from displayed math notation to the next line for the beginning of a paragraph. But I see it happening here all the time, where that explanation seems inapplicable. Why do you and so many others capitalize that initial "W" is the middle of a sentence? ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 16 '14 at 0:30
  • $\begingroup$ @MichaelHardy I think it might just be due to the auto-correct. I don't exactly recall putting a capital there to be honest. $\endgroup$ – Eric Lawson Nov 16 '14 at 0:35
  • $\begingroup$ How does "auto-correct" get involved? I'm typing this on a Google Chrome browser window, and I've never seen anything like that. Does auto-correct come with some browsers or what? ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 16 '14 at 0:37
  • $\begingroup$ @MichaelHardy I am not sure, but while I was typing that comment I typed capital wrong and it auto-corrected it for me. I am on internet explorer so perhaps on that browser it does? $\endgroup$ – Eric Lawson Nov 16 '14 at 0:39
  • $\begingroup$ @MichaelHardy I encounter auto-correct when I work from my ipad. $\endgroup$ – David H Nov 16 '14 at 0:43

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