The tutor shows the motivation behind the integration by parts formula. Integration by parts is a reversal of the product rule (see my post here). If we start with the product rule as (uv)’=uv’ + u’v then we integrate both …
The tutor completes the general solution to the 2nd degree recurrence relation. In my previous post I began seeking the general solution to the recurrence relation tabulated below: n tn 0 1 1 1 2 2 3 3 4 5 …
Math: 2nd degree recurrence relation: final solution Read more »
The tutor starts towards the general solution to a 2nd degree difference equation, aka recurrence relation. Back in my June 21 post, the following recurrence relation emerged: n tn 0 1 1 1 2 2 3 3 4 5 The …
The tutor shows an interesting consequence of mod 3 arithmetic. Back in my March 25, 2014 post, I mentioned that mod means remainder. For example, 19 mod 4 = 3, because when you divide 19 by 4, you get 3 …
The tutor tackles an age-old proof in a new (to him, anyway) manner. A famous theorem of number theory goes like this: For the integers a and b, there exists a solution with integers x and y to ax+by=1 if …
Math: number theory: linear combinations that sum to 1 Read more »
The tutor shows that yesterday’s formulas to generate Pythagorean triples are valid. In yesterday’s post I showed a way to generate Pythagorean triples x, y, z from an odd number n: x n y (n²-1)/2 z (n²+1)/2 Let’s make sure …
Math: Pythagorean triples: proof of yesterday’s generating formulas Read more »
The tutor continues his discussion about Pythagorean triples. Back in my January 7, 2016 post I brought up Pythagorean triples, which are all-integer solutions to x² + y² = z² The equation above is based on the familiar a² + …
Math: number theory: a formula for generating Pythagorean triples Read more »
The tutor discusses the existence of an Euler circuit in a graph. An Euler circuit is a route through a graph that travels every edge exactly once, then ends where it started. If a graph is connected, with every vertex …
The tutor shares a solution to a problem from Grimaldi’s Discrete and Combinatorial Mathematics. Last night I encountered the problem (p. 481) that paraphrases as follows: Imagine a car takes two spaces, a motorbike one. Find a recursive function that …
The tutor points out a handy feature of some Sharp calculators. There are calculators that, once you type the expression and press =, show the answer, but not what you just entered to arrive there. In fact, when I was …
Calculator usage: Sharp’s Multi-Line Playback feature Read more »