Pascal's triangle

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Pascal triangle for n=10.

In mathematics, Pascal's triangle is a representation of the ordered binomial coefficients in the form of a triangle. It is named after the French philosopher and mathematician Blaise Pascal, who introduced this notation in 1654, in his Treatise on the Arithmetic Triangle. Although the properties and applications of the triangle were known to Indian and Chinese mathematicians, Persians, Germans, and Italians before Pascal's triangle, it was Pascal who developed many of its applications and was the first to organize information together.

Pascal's triangle can be generalized to higher dimensions. The three-dimensional version is called Pascal's pyramid or Pascal's tetrahedron, while the more general versions are called Pascal's simplex.^{[citation required]}

History

Chinese arithmetic triangle

The first explicit representation of a triangle of binomial coefficients dates from the 10th century, in the commentaries on the Chandas Shastra, an ancient Indian book of Sanskrit prosody written by Pingala around 200 BCE. C.

The properties of the triangle were discussed by the Persian mathematicians Al-Karaji (953–1029) and Omar Khayyám (1048–1131); hence in Iran it is known as the Khayyam-Pascal triangle or simply the Khayyam triangle. Many related theorems were also known, including the binomial theorem.

In China, this triangle was known since the 11th century by the Chinese mathematician Jia Xian (1010–1070). In the 13th century, Yang Hui (1238–1298) introduced the arithmetic triangle, equivalent to the Pascal's triangle, hence it is called Yang Hui's triangle in China.

Petrus Apianus (1495–1552) published the triangle on the frontispiece of his book on trade calculations Rechnung (1527). This is the first record of the triangle in Europe. In Italy, it is known as the Tartaglia triangle, after the Italian algebraist Niccolò Fontana Tartaglia (1500–1577). It was also studied by Michael Stifel (1486-1567) and by François Viète (1540-1603).

In the Traité du triangle arithmétique (Treatise on the Arithmetic Triangle) published in 1654, Blaise Pascal brings together several well-known results on the triangle, and uses them to solve problems linked to the theory of probability; he proves 19 of his properties, deduced in part from the combinatorial definition of the coefficients. Some of these properties were already known and admitted, but without proof. To prove them, Pascal puts into practice a finished version of mathematical induction. He demonstrates the relationship between the triangle and the binomial formula. It was baptized Pascal's Triangle by Pierre Raymond de Montmort (1708) who called it: Mr. Pascal's Table for Combinations, and by Abraham de Moivre (1730) who called it: "Triangulum Arithmeticum PASCALIANUM" (Latin: "Pascal's Arithmetic Triangle"), which became the modern Western name.

Construction

Pascal's triangle is constructed following a pattern like the one shown in the figure below. It starts from the top with the number «1» downwards (infinity), like a "tree"; it is classified in rows, beginning with row zero (the "1" at the top). This "tree" it has nodes, which are each number that makes up the triangle. If we add two nodes, the node below these two will give us the result, and so on.

The diagonals that start from the «1» located at the head of the triangle are always worth 1.

General use

This triangle was designed to develop the binomial powers. Binomial powers are given by the formula: ${displaystyle (a+b)^{n}}$ where a and b are any constants and n the exponent that defines power. This expression is called Newton's binomial.

This Newton's binomial formula expands the coefficients of each row in Pascal's triangle. This is why there is a close relationship between Pascal's triangle and Newton's binomials.

Each number in the triangle is the sum of the Two. which are located above him.

Link between Pascal's triangle and Newton's binomial

All the figures written in each row of the triangle correspond to the coefficients of the development of the powers of the Newton binomial. A few examples from the series that describe this behavior are:

{displaystyle (a+b)^{2}=1a^{2}+2ab+1b^{2}quad }

{displaystyle (a+b)^{3}=1a^{3}+3a^{2}b+3ab^{2}+1b^{3}quad }

dots

With these examples it can be concluded that the series of the general expression that develops them is:

{displaystyle (a+b)^{n}=a^{n}+a^{n-1}b+a^{n-2}b^{2}...+b^{n}}

Thus, the coefficients developed in the form (a+b)ⁿ are in the row «n+1» of the Triangle of Pascal.

You can also generalize this result for any value ${displaystyle nin mathbb {N} }$ by mathematical induction.

If we call each node of this triangle in each row z, we would be left with the series that describes the general expression of the mode:

{displaystyle (a+b)^{n}=z_{1}a^{n}+z_{2}a^{n-1}b+z_{3}a^{n-2}b^{2}...+z_{n}b^{n}}

In this series ${displaystyle zin mathbb {N} }$ Where z from 1 to n.

Combinatorics in Pascal's Triangle

The construction of the triangle is related to the binomial coefficients according to the formula (also called Pascal ruleCombinative. This formula or rule explains that the coefficients (nodes of the "tree") of a given row of the triangle can be calculated with the combination formula of combinations of combinations of $n$ elements $k$ in $k$ expressed mathematically: ${displaystyle {tbinom {n}{k}}}$ Where $n$ It's the row - 1 and $k$ position in the row.

All this proposal of correlation between combinatorics and Pascal's triangle is given by the aforementioned general rule:

{textstyle (a+b)^{n}=sum _{k=0}^{n}{n choose k}a^{n-k}b^{k}quad {text{para todo}}quad 0leq kleq n;quad n,kin mathbb {N} }

For example, for binomial ${displaystyle (a+b)^{3}}$ We would have the following:

Four nodes would remain (elements composed of a, b and corresponding coefficient) in the developed equation of the binomial, number which refers to the row in which it is found:

{displaystyle (a+b)^{3}=mathbf {1} a^{3}+mathbf {3} a^{2}b+mathbf {3} ab^{2}+mathbf {1} b^{3}}

If we express the coefficients of the triangle of the combination form, the following would be:

{displaystyle {0 choose 0}}

{displaystyle {1 choose 0}quad quad quad {1 choose 1}}

{displaystyle {2 choose 0}quad quad quad {2 choose 1}quad quad quad {2 choose 2}}

{displaystyle {3 choose 0}quad quad quad {3 choose 1}quad quad quad {3 choose 2}quad quad quad {3 choose 3}}

{displaystyle quad leftarrow quad {text{fila correspondiente a }}(a+b)^{3}}

Which corresponding triangle would be:

1

{displaystyle 1quad quad 1}

{displaystyle 1quad quad 2quad quad 1}

{displaystyle 1quad quad 3quad quad 3quad quad 1}

{displaystyle quad leftarrow quad {text{fila correspondiente a }}(a+b)^{3}}

Properties

Pascal triangle with some colored boxes. It can be seen how the values are distributed symmetrically around the vertical axis. The values of the boxes on both sides (in yellow, green and red) have the same value due to the ownership of symmetry

scriptstyle {nchoose {n-p}} = {nchoose p}

. The outer boxes (blue), have null value, and the violet boxes, provide an example of the Pascal rule.

Once the foundations of the intrinsic correlation existing between these two fields of mathematics have been established, see their properties.

This image represents the triangle of Pascal matritally, and also applicable to combination. Each of the values of a Pascal triangle written in a table form correspond to a coefficient of the expansion of a sum power. Specifically, the row number n and column p, corresponds to ${displaystyle {tbinom {n}{p}}}$ or also denoted as ${displaystyle mathbf {C} _{n}^{p}}$ ( ${displaystyle mathbf {C} }$ by "combination") and is said «n on p», «combination of n in p» or "binomial coefficient n, p". The empty boxes correspond to null values (0). Using the properties of the binomial coefficients, you can obtain the following properties of any Pascal triangle with all rigor:

The values of each row of the triangle keep symmetry with respect to the imaginary vertical axis of the same, because ${displaystyle {tbinom {n}{n-p}}={tbinom {n}{p}}.}$

The values corresponding to the area outside the triangle have value 0since ${displaystyle {tbinom {n}{p}}=0}$ When $n}" xmlns="http://www.w3.org/1998/Math/MathML">p▪n{displaystyle ppurn}n}" aria-hidden="true" class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d290fd27ba6202ad4e75b83613783a5821b04b39" style="vertical-align: -0.671ex; margin-left: -0.089ex; width:5.752ex; height:2.176ex;"/>$ .

And of course, the rule of Pascal of constructing the triangle gives the fundamental relationship of the binomial coefficients ${displaystyle {tbinom {n}{p}}+{tbinom {n}{p+1}}={tbinom {n+1}{p+1}}.}$

An interesting consequence of the triangle of Pascal is that the sum of all the values of any row of the triangle is a power of 2. This is because, by the theorem of the binomial, the expansion of the n-Power of ${displaystyle (1+1)^{n}=2^{n}}$ That's it.

{n choose 0} + {n choose 1} + cdots +{n choose n-1} + {n choose n} = 2^n,

which corresponds precisely to the sum of all values in the nth row of a Pascal's triangle.

Approximation to the triangle of Sierpinski obtained colouring in black the boxes of the Pascal Triangle containing a pair and gray number those containing an odd number.

Other interpretations or representations

Right triangle

Pascal triangle in Pascal's original writing.

The illustration at the beginning of the article shows the triangle of Pascal drawn as an equilateral triangle. It is possible to “respect” it in such a way that its drawing is like a right triangle. In this way, on the left is a column of numbers "1". The following column leaves an empty place in the first row and follows with the succession of natural numbers: ${displaystyle 1,2,3,4,dotsn,dots }$ The third column leaves two empty rows and begins with the succession of triangular numbers: ${displaystyle 1,3,6,10,15,dots }$ Drawing this way is easy to see that:

Each number in any column is equal to the partial sum of the elements of the previous column (to the left) to the previous row in descending order.
The third column is the succession of the triangular numbers; the fourth, that of the tetraherdrical numbers; the fifth, that of the pentaedrical numbers, and so on.

Powers in base 2

You can also find the power base 2 of the form ${displaystyle 2^{n}, nin mathbb {N} }$ as the successive sums of the coefficients of the rows, being n the row in which the power is found $2^{n}$ :

{displaystyle 2^{0}=1}

{displaystyle 2^{1}=1+1}

{displaystyle 2^{2}=1+2+1}

{displaystyle 2^{3}=1+3+3+1}

{displaystyle 2^{4}=1+4+6+4+1}

dots

Fibonacci Sequence

Relationship between the triangle of Pascal and the succession of Fibonacci

In the triangle of Pascal you can appreciate a relationship between a mode of adding the diagonals and the succession of Fibonacci. The first terms of this succession are: ${displaystyle 1,1,2,3,5,8,13,21,34,55dots }$

As can be seen in the image on the right, the successive sums of the diagonals from top right to bottom left make up the Fibonacci sequence.

Prime numbers

There is a property about Pascal's triangle that if the first element in a row (not counting the 1's) is a prime number, then all other elements in the row will be divisible by it.

Example:

${displaystyle {text{En la fila 11 }}rightarrow 1quad 11quad 55quad 165quad 330quad 462quad 462quad 330quad 165quad 55quad 1}$ ;

The 55, 165, 330 and 462 are all divisible by 11.

Generalizations

Combined example of trinomial coefficient.

Instead of considering the powers of a + b, one can consider those of the trinomial a + b > + c. Thus, (a + b + c)ⁿ is a sum of monomials of the form λ_{p, q, r} a^pb^qc^r, with p, q, and r positive, p + q + r = n, and λ_{p, q, r} a natural number called the trinomial coefficient. The calculations are similar to those of the binomial coefficient, and are given by the following expression:

lambda_{p,q,r} = {n choose p,q,r} = frac {n!} {p! q! r!}

in subsets of p, q, and r elements.

Pascal pyramid. The first sections have been drawn from the summit.

These coefficients can be thought of as the three-dimensional analogy of Pascal's triangle. In fact, the distribution of these coefficients in a pyramidal style is known as Pascal's pyramid; is also infinite, with triangular sections, and the value in each box is the sum of the values in the three boxes above it.

This pyramid is observed to be invariant by rotation of 120 degrees around a vertical axis that passes through the vertex. Pascal's triangle appears on all three faces of the pyramid.

Similarly, all this can be generalized to any finite dimensions, but without the possibility of making simple explanatory pictures.

Pascal's triangle in programming

The following codes of the triangle of Pascal are based on the property: ${displaystyle n=C(fila,posicion)}$ taking into account that n is a value belonging to the triangle of Pascal, row is the level number (height) that we find of the triangle of Pascal, would begin by "1", which would be row 1, and the position refers to the index of the data in the row, read from left to right. Thus, in the second row of the triangle of Pascal we have: ${displaystyle 2=C(2,1)}$ as it is in position 1 (start by position 0) and is at level 2. Bearing in mind that C refers to the combination, so the following formula is followed to find it: ${displaystyle C(x,y)={x! over y!(x-y)!}}$ . And considering it is made use of the factorial.

On the other hand, in the Java code the factorial is performed by a loop, following the definition: ${displaystyle x!=x(x-1)...1}$

On the other hand, the Python code is performed through recursivity, following the following definition: ${displaystyle x!=x(x-1)!}$

Code in Java

import java.util.Scanner;public class TrianguloPascal {public static void main(String[] args) {Scanner Entry = new Scanner(System.in);System.out.print("Indicate the number of rows you want: ");create Triangulo(Entry.nextInt());Entry.close.();!private static long factorial(int n1) {long miNum = 1;for(int i=1;i♫n1;i+) {miNum♪i;!return miNum;!private static int combination(int n1, int n2) {int result = 0;result = (int) (factorial(n1) / (factorial(n2)♪factorial(n1-n2));return result;!public static void create Triangulo(int No.) {for(int row=1;row♫No.;row+) {for(int i=0;i.(No.-row);i+) {System.out.print(");!if(row ♪ 1) {System.out.println("1 1");!else {for(int i=0;i.(row+1);i+) {System.out.print(combination(row, i) + ");!System.out.println();!!!!

Code in Python

# Doing the Pascal's Triangle using Pythondef factorial(num(c): if num ▪ 0: # Doing the factorial using recursion return int(num♪factorial(num-1) else: return 1def combination(num1, num2(c): return int(factorial(num1) / (factorial(num2)♪factorial(num1-num2())))def create Triangulo(No.(c): for row in Range(No.(c): for j in Range(No.-row+1(c): print(", end=") if row ♪ 0: print("1 1") else: for j in Range(row+2(c): print(combination(row+1, j), end=") print()create Triangulo(int(input("Indicate the number of rows you want: "())))

Code in C++

# Include▪using namespace std;long factorial(int x♪if(x ▪ 0♪return x ♪ factorial(x-1); //Recursivity!else{return 1;!!int combination(int num1, int num2♪return (int) factorial(num1)/ (factorial(num2)♪factorial(num1-num2));!void PrintTriangulo(int rows♪for(int i=1;i♫rows;i+♪for(int j=0;j.(rows-i);j+♪cout;";!if(i ♪ 1♪cout;"1 1";endl;!else{for(int j=0;j♫i;j+♪cout;combination(i, j); ";!cout ;endl;!!!int main(){int rows = 0;cout;"Write the number of levels you want: ";endl;std::cin ◊ rows;PrintTriangulo(rows); return 0;!

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