Let hn denote the number of regions into which a convex polygonal region with n+2 sides is divid

Let hn denote the number of regions into which a convex...

Key Concepts

Recurrence Relations

Recurrence relations are equations that recursively define a sequence, where each term in the sequence is defined as a function of the preceding term(s). They are widely used in combinatorics to break complex counting problems into simpler, more manageable subproblems by establishing relationships between different instances of the problem.

Generating Functions

Generating functions are formal power series in which the coefficients encode information about a sequence. They provide a powerful tool for solving recurrence relations and combinatorial problems by transforming sequences into algebraic expressions, enabling techniques such as finding closed-form solutions and extracting combinatorial quantities.

Binomial Coefficients

Binomial coefficients, often denoted as combinations, count the number of ways to select a subset of elements from a larger set. They are fundamental in combinatorial analysis and appear in various contexts, including the expansion of generating functions, and in counting arguments that involve choosing particular subsets of elements.

Closed-Form Solutions

A closed-form solution is an explicit expression that directly calculates the nth term of a sequence without requiring recursion. In combinatorics, deriving a closed-form solution from a recurrence or generating function allows for direct computation and deeper insights into the asymptotic behavior and combinatorial structure of the sequence.

Transcript

00:01 So in this question, we're tasked to prove that for any polygon with a number of sides greater than 3, so 4 is the minimum, the number of diagonals that can be drawn is equal to n times n minus 3 divided by 2, where n is the number of sides that the polygon has.

00:22 So to define a diagonal, it's a line that can be drawn from a vertex of a polygon to any other vertex that is not already connected with that vertex by a side.

00:37 So this side and this side are already connected by, or this vertex and this vertex are already connected by a side of the polygon.

00:47 So drawing a line between those two vertices does not form a diagonal.

00:52 So in order to prove that this is true, we need to use the process of proof by mathematical induction.

01:05 And that involves three steps, proving the base case.

01:08 So proving that for the least number of, or for the polygon with the least number of sides, the statement is true.

01:18 Forming the inductive hypothesis.

01:20 So for any integer side length or side number.

01:27 Or k, the statement is true, or the statement is assumed true.

01:32 And then finally, step three, proving the inductive case, which is to prove that, or which is to prove that for any k plus one, where k is the integer from step two, the statement also holds.

01:47 So by doing this, by proving that the statement is true for the smallest possible integer, in this case, that is four.

01:59 And for every integer that makes the statement true, the following integer also makes it true.

02:07 We've proven that every integer makes this true after 4.

02:13 So let's start with the base case, which is a four -sided polygon or a quadrilateral.

02:19 So the way that i do this is by going from vertex to vertex in a clockwise order and drawing drawing diagonals from there.

02:32 So, drawing diagonals from there.

02:35 So let's start with this one.

02:37 So here we can only draw one diagonal, and let's label this vertex with the one for the number of diagonals that can be drawn, and let's move to this one.

02:50 This one also has one diagonal that can be drawn, and then for every successive one, there are no diagonals that can be drawn.

03:03 So, n is equal to four in this case.

03:06 So, must be equal to 4 times 4 minus 3 divided by 2, which is equal to 4 times 1 over 2, which is equal to 4 over 2, which is equal to 2.

03:20 So when n equals 4, d equals 2.

03:24 And that proves the statement is true for the base case.

03:29 So for the inductive case, or to form the inductive hypothesis, we need to say that for any integer, k, greater than 3, the statement is true.

03:49 So the statement is d equal to k times k minus 3 over 2.

03:58 So this statement right here is assumed true for now.

04:04 Finally, we're at step 3, which is to prove the inductive case.

04:08 That is to prove that for any integer k plus 1, the statement is true, where n is equal to k plus 1.

04:22 So first, let's look at the process itself in which we draw diagonals.

04:29 So i'm going to take a pentagon here, for example, where n equals 5.

04:34 And let's take a look step by step how everything works.

04:38 So let's start with this vertex right here.

04:43 So we can draw two diagonals.

04:48 And the reason that we can draw only two diagonals when n is equal to 5 from this one vertex, from this first vertex, is because.

04:56 Out of the five vertices that exist in this polygon, one is the vertex itself, so that subtracts from one from the number of vertices available.

05:10 And the two adjacent vertices are already connected to that vertex by side lengths, so that's two more vertices that are made unavailable.

05:20 So for any number and or any integer and greater than four, for the first vertex, there are only n -3 diagonals that can be drawn because three of the vertices are ineligible for vertices for diagonals from that vertex.

05:48 So again, i'm going to label each vertex with the number of diagonals that can be drawn from that point.

05:57 So the first vertex will be labeled n -3.

06:01 I'm going to use n in this case because it's more high.

06:03 Helpful.

06:04 So for the following vertex, it's actually the same case.

06:09 The first vertex is already connected to that vertex by a side length, or by a side, so it's ineligible.

06:20 So there are only two vertices again...