Another long one. Obviously not for the faint of heart :) Anyway, this one is about the beautiful discrete counterpart of calculus, the calculus of sequences or the calculus of differences. Pretty much like in Alice's Wonderland things are strangely familiar and yet very different in this alternate reality calculus. Featuring the Newton-Gregory interpolation formula, a powerful what comes next oracle, and some very off-the-beaten track spottings of some all-time favourites such as the Fibonacci sequence, Pascal's triangle and Maclaurin series. 00:00 Intro 05:16 Derivative = difference 08:37 What's the difference 16:03 The Master formula 18:19 What's next is silly 22:05 Gregory Newton works for everything 28:15 Integral = Sum 32:52 Differential equation = Difference equation 36:06 Summary and real world application 39:22 Proof Here is a very nice write-up by David Gleich with a particular focus on the use of falling powers. https://tinyurl.com/ymcyrapz This is a nice lesson from a Coursera course on this topic https://www.coursera.org/lecture/discrete-calculus/differences-k4jBq One volume of Schaum's outlines is dedicated to "The calculus of finite differences and difference equations" (by Murray R. Spiegel) Examples galore! This is a really nice very old book Calculus Of Finite Differences by George Boole (published in 1860!) https://tinyurl.com/3bdjr932 Thanks to Ian Robertson for recommending this one. There is a wiki page about our mystery sequence: https://tinyurl.com/uwc89yub It's got a proof for why the mystery sequence counts the maximal numbers of regions cut by those cutting lines. If you have access to the book "The book of Numbers" by John Conway and Richard Guy, it's got the best proof I am aware of. Here is a sketch of how you solve the Fibonacci difference equation to find Binet's formula https://imgur.com/a/Btu5ZVk Here are a couple more beautiful gems that I did not get around to mentioning: 1. When we evaluate the G-N formula for 2^n what we are really doing is adding the entries in the nth row of Pascal's triangle (which starts with a 0th row :) And, of course, adding these entries really gives 2^n. 2. Evaluating the G-F formula for 2^n at n= -1 gives 1-1+1-1+... which diverges but whose Cesaro sum is 2^(-1)=1/2!! Something similar happens for n=-2. 3. In the proof at the end we also show that the difference of n choose m is n choose m-1. This implies immediately that the difference of the mth falling power is m times the difference of the m-1st falling power. Today's music is by "I promise" by Ian Post. Enjoy! Burkard P.S.: Some typos and bloopers https://www.youtube.com/watch?v=4AuV93LOPcE (396 should be 369) https://www.youtube.com/watch?v=4AuV93LOPcE (where did the 5 go?) https://www.youtube.com/watch?v=4AuV93LOPcE (a new kind of math(s) :)