VIEWS: 2 PAGES: 12 POSTED ON: 3/16/2011
History This triangle appears in Treatise on the Arithmetic Triangle, by Blaise Pascal. It was written in 1653 and published posthumously in 1665. Pascal was one of the founders of probability theory. He is credited with inventing the syringe, wheelbarrow, hydrolic press, and a calculating machine (our first calculator). Pascal gave up doing mathematics to pursue religion rather rigorously, and he wrote many philosophical papers on the subject. However, he returned to mathematics at the end of his life when he got a terrible toothache, which he perceived as a sign from god that he should do mathematics again. In actuality, the triangle was not invented (discovered) by Pascal. It was known in India around 200 BCE, and appears in many medieval Islamic mathematics texts. Similar arrangements of numbers were found in the works of Persian mathematicians Al-Karaji and Omar Khayyam. The triangle appeared in Europe in 1225 in Jordanus de Nomore’s On Arithmetic. In 1556 the triangle appeared in Niccolo of Brescia’s General Treatise. Niccolo is more commonly know as Tartaglia the stammerer, due to an injury he sustained as a boy. In Italy, the triangle is known as Tartaglia’s triangle. Pascal’s name was attached to the triangle in 1708 by Pierre Remond de Montmort. Combinations n We need to recall what C (n, r ) means. This is equal to the number of r combinations of n objects taken r at a time. In other words it is the number of ways of choosing r objects out of a possible n objects if we don’t care about what order they are chosen. We studied this concept in section 6.4. Here we developed the formula: n n! C (n, r ) r r !(n r )! n These things are often called Binomial Coefficients, for reasons that will be apparent r shortly. Binomial Theorem Consider the following: (a b) a 2 2ab b 2 2 (a b)3 a3 3a 2b 3ab 2 b3 (a b)4 a 4 4a3b 6a 2b 2 4ab3 b3 (a b)5 a5 5a 4b 10a 3b 2 10a 2b3 4ab 4 b5 Do you see the pattern? The entries in the triangle appear as the coefficients in the expansion. More generally we have: n n n n 1 n1 n 0 n n n nr r (a b)n a nb0 a n1b1 a n2b2 a b a b a b . 0 1 2 n 1 n r 0 r Thinking of the coefficients as combinations, is there any way we can justify this phenomenon? Let’s think about it. Perhaps it would be best to multiply (a b)3 very slowly, without combining terms. This is called multiplying it formally: (a b)3 aaa aab aba baa abb bab bba bbb Notice that there are a total of 8 terms appearing on the right hand side of this equation. Now what we want to do is combine like terms. A term is like another if it has the same number of a’s and b’s in it. Order doesn’t matter. So for example, all of the terms with two a’s and one b are: aab, aba, baa. There are three of these in this case. This is because out of all possible three 3 letter “words” using only the letters a and b, there are 3 with exactly 1 b. We will see a 1 a b aaaa 4 aaab aaba abaa baaa similar result if we do: aabb abba bbaa abab baba baab abbb babb bbab bbba bbbb Notice that there are a total of 16 terms appearing on the right hand side of this equation. Now when we go to combine like terms we are thinking of a term as a “word” with 4 letter words 4 using only the letters a and b. So, for example there are 6 words with 2 a’s and 2 b’s, 2 which is exactly what the coefficient of the a 2b2 term. In general when we are expanding (a b) n we formally get a whole bunch of n letter “words” using only the letters a and b. Finding the number of terms which combine to give a nk bk is equivalent to finding the number of n letter “words” with a and b as letters which have n exactly k b’s. This is exactly what gives you since you can think of looking at n slots and k choosing the k slots that the b’s will go in. All of this stuff is often called the Binomial Theorem basically because it gives us a way to expand the binomial (a b) n . This is also why these combinations are often called binomial coefficients. What is cool is that, if you have Pascal’s triangle handy, you can fairly quickly multiply out a binomial. Binomial Theorem Problems: 1. Use Pascal’s triangle to expand the following: (a). (2 x 3) 6 (b). ( x 2 y )5 2. Expand Plug in x 1 (1 10 x) = 1 (1 10 x) 2 = (1 10 x)3 = (1 10 x) 4 = (1 10 x)5 = Now plug in x 1 into both sides of the equations above. What do you see? What happened with the last one? 3. Use Pascal’s triangle to find the coefficient of the given term in the given expansion: (a). (2 x)14 ; x 9 100 1 x91 (b). x ; 2 2 x 3 12 (c). ; x7 4. First expand using Pascal’s Triangle: ( x y)n = Now plug in x 1 and y 1 into both sides. What do you see? Identities Now that we have this triangle we can look at it and find out some interesting facts about these binomial coefficients. Here is a list of a few of them: n n 10 10 2356 2356 for example or k nk 7 3 356 2000 n 1 n 1 n 7 7 8 123 123 124 for example or k k 1 k 3 2 3 47 46 47 n n n n n 6 6 6 6 6 6 6 2 for example 2 n 6 0 1 2 n 1 n 0 1 2 3 4 5 6 n n 1 n 2 n k n k 1 5 6 7 8 9 10 for example 0 1 2 k k 0 1 2 3 4 5 k k 1 k 2 n n 1 2 3 4 5 6 7 8 for example k k k k k 1 2 2 2 2 2 2 3 Try checking a few of these on the triangle. See if you can come up with some way of justifying why they are true. The first of these simply says that we should see some symmetry in the triangle. And indeed we do. All of the numbers are symmetric around the center of the triangle. Why? The second of these simply expresses what we did to make the triangle in the first place. We add up the two consecutive numbers in the row above. We can see that the third of these is true by simply adding up each row of the triangle: 1 =1=20 1 + 1 =2=21 1 + 2 + 1 =4=22 1 + 3 + 3 + 1 =8=23 1 + 4 + 6 + 4 + 1 =16=24 If you look at the last problem in the last section you will see a proof of this fact. Can you give another? Binomial Distribution Triangular Numbers and Other Figurate Numbers Heading down the triangle along a south-east diagonal we see the following sequences of numbers appearing: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 … 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 … 1, 3, 6, 10, 15, 21, 28, 36, 45, 55 … 1, 4, 10, 20, 35, 56, 84, 120, 156, 211 … 1, 5, 15, 35, 70, 126, 210, 330, 486, 697 … Etc. The first of these is just a sequence of all 1’s. Not terribly interesting. The second of these is obviously the sequence of counting numbers, sometimes called the Natural Numbers. The third of these are known as the Triangular Numbers. This is because they can be arranged in the following way: 1 3 6 10 15 The fourth of these are known at the Tetrahedral Numbers. The picture for these is very similar. Except now we have moved into three dimensions. Think of stacking cannon balls into triangular pyramids. The number of balls you use is a tetrahedral number: 4 20 20 Can you think of some way of visualizing the next set of numbers or the one after that? Catalan Numbers Consider the following problem: How many ways are there for an even number of people all seated around a circular table to shake hands if no arms are allowed to cross? Everyone has to shake the hand of somebody and no one can shake their own hand. If we think of these people as points around a circle, and we think of shaking hands as drawing a line between two points, then we are looking for all of the ways of drawing such lines so that no two of them cross: For 2 People there is only 1 way. For 4 People there are 2 ways. For 6 People there are 5 ways What about for 8 People? Can you find all of the possible configurations? The number of ways of arranging that 2n people around a table shake hands without their arms crossing is called the nth Catalan Number, Cn. The first few Catalan numbers are: 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, ... Note that C1 =1, C2 =2, C3 =5, just like we found in our pictures. The Catalan numbers actually appear in the triangle but they are kind of hidden. Can you find them? Look down the center column of the triangle. This column reads: 1, 2, 6, 20, 70, 252, 924, 3432 Can you see them yet? Try dividing these numbers progressively by the natural numbers: 1 2 6 20 70 252 924 3432 , , , , , , , , 1 2 3 4 5 6 7 8 You end up getting: 1, 1, 2, 5, 14, 42, 132, 429, … !!!!!! So if we want to get a formula for the nth Catalan number we can just use our formula for the elements in the triangle as binomial coefficients. The elements of the center column all look 2n like , and what we did in each case was just divide them each by n+1. Thus we get the n 1 2n formula: Cn n 1 n Serpinski’s Triangle Idea: Look at the remainders in the triangle when you divide each of the entries by various fixed numbers. Color the different remainders different colors and see what happens. Fibonacci Numbers Idea: Move along a diagonal path adding values in the triangle and you will see the Fibonacci Sequence. There are various other things that can be done with the Fibonacci Numbers as well. Reappearance of Numbers Idea: Some numbers seem to reappear in the triangle: Eg. 120 appears twice in the 10th row, the 16th row, and the 120th row. 210 appears twice in the 10th, 20th, and 210th row. 3003 appears twice in the 14th, 15th, 78th, and 3003rd row. It is known that there are infinitely many numbers that occur at least 6 times in Pascal's triangle, namely the solutions to given by where Fi is the ith Fibonacci number (Singmaster 1975). The first few such values of r for k = 1, 2, ... are 1, 3003, 61218182743304701891431482520, .... Points on a Circle Image Points Segments Triangles Quadrilaterals Pentagons Hexagons Heptagons 1 2 1 3 3 1 4 6 4 1 5 10 10 5 1 6 15 20 15 6 1 7 21 35 35 21 7 1 As you may have noticed, the numbers in the chart above are actually the numbers in Pascal's Triangle, except the first 1 in each row is missing. The circular figures are formed by simply placing a number of points on a circle equally spread apart, and then drawing all the possible lines between them. This chart shows that for a figure with n points, all you need to do is look at the nth row of the Pascal’s triangle in order to find the number of points, line segments, and polygons in the figure. Note that the polygons we are considering are only those which have all of their vertices on the circle.