Figurate and Polygonal NumbersDate: 11/21/98 at 18:24:55 From: Megan Subject: Figurate numbers I need to know everything about figurate numbers - no one here knows anything. Date: 11/22/98 at 15:48:54 From: Doctor Floor Subject: Re: Figurate numbers Hi Megan, Thank you for your question. Figurate numbers are number sequences that are found by creating consecutive geometrical figures. Most well known are so-called polygonal numbers, and most well known among these are squares. Polygonal numbers are found by making consecutive nested polygons. When we take squares as an example, they can be created this way: 1 1 2 1 2 3 (the numbers should be thought of as numbered dots) 2 2 2 2 3 3 3 3 The first square has 1 dot, so the first square number is 1. The second square has 4 dots, so the second square number is 4. And the third is 9. Of course you know the formula for square numbers is f(n)=n^2. For bigger polygons the type of nesting must be considered carefully. I will give pentagonal numbers as an example: Pentagonal numbers are the numbers of dots you need for consecutive nested pentagons (figures with five sides). Of course the first pentagon is 1 dot, so the first pentagonal number is 1. The second pentagon has sides of two dots. The first pentagon must be included: 2 2 1 2 2 So the second pentagonal number is 5. The third pentagon has sides of three dots. For the third pentagonal number, the first two pentagons must be nested inside this one (the dots with 3 are added): 3 3 3 2 2 1 3 2 2 3 3 3 So the third pentagonal number is 12. As a final example I will give the fourth pentagonal number. First create the figure: 4 4 4 4 3 3 3 2 2 4 1 3 2 2 4 3 3 3 4 4 4 4 We see that the fourth pentagonal number is 22. So we see that the polygons are nested in one vertex. This is true for all polygonal numbers. Other types of figurate numbers are acquired by other types of nesting, but I don't know about them. We can make a general formula that gives the rth n-gonal number (where n is the number of vertices in the polygon): n-2 2 n-4 --- r - --- r 2 2 or 0.5r[(n-2)r-(n-4)] To derive the formula, first note that it must be quadratic. Then consider the 0th, 1st and 2nd n-gonal numbers for some n. They are very easy: 0 1 n Now say the formula for the rth n-gonal number is F_n(r) = ar^2 + br + c. Since F_n(0) = 0 it follows that c = 0. F_n(1) = 1 gives a+b = 1, or a = 1-b [1]. F_n(2) = n gives 4a+2b = n [2]. Inserting [1] in [2] gives 4(1-b)+2b = n, so -2b = n-4 and b = -0.5(n-4) [3]. Inserting [3] in [1] gives a = 1+0.5(n-4) = 0.5(n-2). And we find the formula f_n(r) = 0.5r[(n-2)r-(n-4)] as given above. I hope this helps! Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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