Dr. Claus Volko

Some months ago I applied for membership in Intertel. There was some technical problem with the website and I decided to abandon the venture, as my interest was not that great anyway. A couple of days ago I received an email from an Intertel officer, which eventually had the effect that I decided to join Intertel after all. In contrast to most of the other high IQ societies I am a member of, Intertel membership is not free of charge. The annual fee is USD 35. I also had to fill in a form in which I was asked about, among other things, my educational level and my religious views. That reminded me that after all, formal education does play a role in our society and it was not totally in vain that I completed a doctorate. And yet, this is old-fashioned. Society has changed very much in the past few years. In the groups at Facebook where I occasionally spend time discussing about things that interest me, nobody cares about degrees. What matters is whether what you write makes sense.

Beim Aufräumen dieses Blogs habe ich leider auch ein wertvolles Posting über Endliche Differenzen gelöscht. Dieses möchte ich nun wieder herstellen. Ende 2010 habe ich die Beobachtung gemacht, dass sich Summen auf eine bestimmte Weise darstellen lassen. Ich habe darüber hier geschrieben:  https://www.pouet.net/topic.php?which=7754 Daraufhin hat mich ein Mitlesender auf das folgende Paper verwiesen:  https://www.cs.purdue.edu/homes/dgleich/publications/Gleich%202005%20-%20finite%20calculus.pdf Es gilt: x^ m = x (x - 1) (x - 2) ... (x - (m - 1)) ∆x^ m  = (x + 1)^ m  - x^ m  = (x + 1) x (x - 1) ... (x + 1 - (m - 1)) - x^ m  = m x^( m - 1) Daraus folgt zum Beispiel: ∆x^ 2  = (x + 1) x - x (x - 1) = x^2 + x - x^2 + x = 2x ∆x^ 3  = (x + 1) x (x - 1) - x (x - 1) (x - 2) = 3x^ 2  = 3x (x - 1) => ∑x (x - 1) = ∑x^ 2  = 1/3 ∑∆x^ 3  = 1/3 (∑(x = 1 to n + 1)∆x^ 3  - ∑(x = 1 to n)∆x^ 3 ) = 1/3 (n+1)^ 3  = 1/3 (n + 2) (n + 1) n => ∑x (x + 1) = 1/3 (n + 1) n (n - 1) (Wenn nicht