This may look like a neatly arranged stack of numbers, but it's actually a mathematical treasure trove. Indian mathematicians called it the Staircase of Mount Meru. In Iran, it's the Khayyam Triangle. And in China, it's Yang Hui's Triangle. To much of the Western world, it's known as Pascal's Triangle after French mathematician Blaise Pascal, which seems a bit unfair since he was clearly late to the party, but he still had a lot to contribute. So what is it about this that has so intrigued mathematicians the world over? In short, it's full of patterns and secrets. First and foremost, there's the pattern that generates it. Start with one and imagine invisible zeros on either side of it. Add them together in pairs, and you'll generate the next row. Now, do that again and again. Keep going and you'll wind up with something like this, though really Pascal's Triangle goes on infinitely. Now, each row corresponds to what's called the coefficients of a binomial expansion of the form (x+y)^n, where n is the number of the row, and we start counting from zero. So if you make n=2 and expand it, you get (x^2) + 2xy + (y^2). The coefficients, or numbers in front of the variables, are the same as the numbers in that row of Pascal's Triangle. You'll see the same thing with n=3, which expands to this. So the triangle is a quick and easy way to look up all of these coefficients. But there's much more. For example, add up the numbers in each row, and you'll get successive powers of two. Or in a given row, treat each number as part of a decimal expansion. In other words, row two is (1x1) + (2x10) + (1x100). You get 121, which is 11^2. And take a look at what happens when you do the same thing to row six. It adds up to 1,771,561, which is 11^6, and so on. There are also geometric applications. Look at the diagonals. The first two aren't very interesting: all ones, and then the positive integers, also known as natural numbers. But the numbers in the next diagonal are called the triangular numbers because if you take that many dots, you can stack them into equilateral triangles. The next diagonal has the tetrahedral numbers because similarly, you can stack that many spheres into tetrahedra. Or how about this: shade in all of the odd numbers. It doesn't look like much when the triangle's small, but if you add thousands of rows, you get a fractal known as Sierpinski's Triangle. This triangle isn't just a mathematical work of art. It's also quite useful, especially when it comes to probability and calculations in the domain of combinatorics. Say you want to have five children, and would like to know the probability of having your dream family of three girls and two boys. In the binomial expansion, that corresponds to girl plus boy to the fifth power. So we look at the row five, where the first number corresponds to five girls, and the last corresponds to five boys. The third number is what we're looking for. Ten out of the sum of all the possibilities in the row. so 10/32, or 31.25%. Or, if you're randomly picking a five-player basketball team out of a group of twelve friends, how many possible groups of five are there? In combinatoric terms, this problem would be phrased as twelve choose five, and could be calculated with this formula, or you could just look at the sixth element of row twelve on the triangle and get your answer. The patterns in Pascal's Triangle are a testament to the elegantly interwoven fabric of mathematics. And it's still revealing fresh secrets to this day. For example, mathematicians recently discovered a way to expand it to these kinds of polynomials. What might we find next? Well, that's up to you.
Ovo možda izgleda samo kao uredno složen stog brojeva, ali zapravo je matematičko skriveno blago. Indijski matematičari zvali su ga Stepenice planine Meru. U Iranu, to je Khayyamov trokut, a u Kini, Yang Huijev trokut. Većini zapadnog svijeta poznat je kao Pascalov trokut po francukom matematičaru Blaiseu Pascalu, što je ipak malo nepravedno, jer očito nije izmislio ništa novo, ali ipak je i on dao svoj doprinos. Ali zašto je toliko zaokupljao matematičare diljem svijeta? Ukratko, pun je obrazaca i tajni. Prvo i najvažnije, uzorak koji ga generira. Počnite s jedinicom i zamislite nevidljive nule s obje strane jedinice. Zbrojite po dva broja, i generirat ćete slijedeći red. Sada ponavljajte postupak. Nastavite i dobit ćete nešto poput ovog, iako se Pascalov trokut nastavlja u beskonačnost. Svaki red odgovara nečemu naziva binomni koeficijenti raspisa (x+y)^n, gdje je n broj reda, ako krećemo brojiti od nule. Na primjer ako raspišemo izraz za n=2, dobit ćemo (x^2)+2xy+(y^2). Koeficijenti, ili brojevi ispred varijabli, jednaki su brojevima u odgovarajućem redu Pascalovog trokuta. Isto možete vidjeti i za n=3, što se raspisuje ovako. Trokut je dakle brz i jednostavan način za pronalaženje koeficijenata. Ali to nije sve. Na primjer, zbrojite brojeve u svakom redu, i dobit ćete uzastopne potencije od dva. Ili u bilo kojem redu, gledajte svaki broj kao dio decimalnog zapisa. Drugim riječima, drugi red je (1x1) + (2x10) + (1x100). Rješenje je 121, što je 11^2. Pogledajte što će se dogoditi kada napravite isto u šestom redu. Rješenje je 1 771 561, što je 11^6, i tako dalje. Postoje i geometrijske primjene. Pogledajte dijagonale. Prve dvije nisu posebno zanimljive: samo jedinice, a zatim pozitivni brojevi; poznatiji kao prirodni brojevi. Ali brojevi u slijedećoj dijagonali zovu se trokutasti brojevi jer ako uzmete toliko točkica, možete ih naslagati u jednakostranične trokute. Slijedeća dijagonala ima tetraedne brojeve jer se navedeni broj sfera može naslagati u tetraedar. Ili primjerice: zasjenčajte sve neparne brojeve. To nije posebno zanimljivo kada je trokut mali, ali ako se dodaje tisuće redova, dobije se Sierpinskijev fraktal. Ovaj trokut nije samo matematičko umjetničko djelo. On je i koristan, posebice u područjima vjerojatnosti i računanju u području kombinatorike. Ako primjerice želite imati petero djece i želite znati koja je vjerojatnost da dobijete kako ste sanjali: tri djevojčice i dva dječaka. U zapisu pomoću potencije binoma, to odgovara djevojčici + dječaku na petu potenciju. Pa pogledajmo peti red, gdje prvi broj odgovara pet djevojčica, a posljednji pet dječaka. Mi tražimo treći broj. Deset kroz zbroj svih mogućnosti u tom redu. pa je to, 10/32, ili 31.25%. Ili, ako nasumično izabirete peteročlanu košarkašku momčad iz skupine od 12 prijatelja, koliko mogućih grupa od pet osoba postoji? Jezikom kombinatorike, ovaj problem izražen je kao 12 povrh 5, i računa se pomoću ove formule, ili jednostavno možete pogledati šesti element dvanaestog reda u trokutu i dobit ćete odgovor. Obrasci u Pascalovom trokutu dokaz su elegantnog ispreplitanja djelova matematike. Sve njegove tajne još nisu otkrivene. Na primjer, matematičari su nedavno otkrili način kako ga proširiti na ovu vrstu polinoma. Što ćemo naći slijedeće? To je na vama.