Triangle of Pascal

See also: Pascal

In Mathematical, the triangle of Pascal , is a geometrical arrangement of the binomial coefficients in a Triangle. At the line I and the column J (0 ≤ J ≤ I ) is placed the binomial coefficient $\{I\; \backslash \; choose\; J\}$.

It was very early used to develop expressions of the form $(has\; +\; b)^n\; \backslash ,$

History

The tradition allots the name of triangle of Pascal to the triangle describes higher. However, this triangle was already known in the East and Middle-East several centuries before the publication of Blaise Pascal. He was thus known Persan mathematicians, for example Al-Karaji (953 - 1029) or Omar Khayyam at the 11th century which uses it to develop (has + b)ⁿ. He appears in China since 1261 in a work of Yang Hui (with row 6) and in the Miroir of jade of the four elements of Zhu Shijie in 1303 (with row 8). Yang Hui allots the paternity of the triangle to the Chinese mathematician of XIe century Jia Xian. This triangle made it possible to present the coefficients of the various terms in the formula of the binomial and, according to V.J. Katz, it was used to generalize with degrees higher than two the method of extraction of root.

In Europe, it appears in the work of Peter Apian, Rechnung (1527). He is studied by Michael Stifel (1486 - 1567) and Tartaglia (1499 - 1557). It is besides under the name of Triangle of Tartaglia that it is known in Italy. But it is Blaise Pascal which devotes a treaty to him: the Treated arithmetic triangle (1654) showing 19 of its properties, properties rising partly from the combinative definition of the coefficients. Many these properties already known but allowed and were not shown. To show them, Pascal sets up in his treaty a succeeded version of the Raisonnement by recurrence. He shows there the bond between the triangle and the formula of the binomial. He exploits it in a problem of launching balanced part (problem of the parties).

Construction

Combinative

By writing the formula of Pascal ,

for all entireties I and J such as 0 < J < I , $\{I\; \backslash \; choose\; J\}\; =\; \{i-1\; \backslash \; choose\; j-1\}\; +\; \{i-1\; \backslash \; choose\; J\}$

we notice that the coefficient of the line I and column J is obtained by adding the coefficients of the line I - 1 and column J - 1 and of the line I - 1 and column J . Moreover we know that

$\{N\; \backslash \; choose\; 0\}\; =\; \{N\; \backslash \; choose\; N\}\; =1$.

We deduce from it a method of construction of the triangle of Pascal:

• we place in column 0 of the 1 at each line, and of the 1 at each entry of the diagonal,
• on the basis of the top and going down, we supplement the triangle by adding two adjacent coefficients of a line, to produce the coefficient of the lower line, in lower part of the coefficient of right-hand side.

According to the following diagram, it is simple not to be mistaken:

WITH + B || (A+B)

Matric

Easy to build starting from the Factorial S, it is possible to represent the triangle of Pascal using the Exponentielle of a matrix: the triangle is the result of exponential of a matrix whose diagonal contains 1,2,3,4, … , zero elsewhere.

Data processing

Let us write the algorithm, in formal language, of construction of the triangle of Pascal. Note that this algorithm creates a new line starting from the preceding one.

Variables: Table of 1 to X of table of 1 to X of entireties C (two-dimensional table) Entireties I , J , N , X

N ← 10 C [0] ← 1 for I of 1 with N to make C [0] ← 1 C [I] ← 1 for J of 1 with I -1 to make C [ J ] ← C [ J -1] + C [ J ] finpour finpour afficher_tableau ( C )

Properties

Dependant on construction

• the sum of the terms of a line: the sum of the terms on the line of row N (first line = row 0) is equal to 2ⁿ.
• hockey sticks: If one makes the sum of the terms, on the basis of an edge of the triangle and going down vertically, one obtains the term located in diagonal in bottom on the right of the last term of the column. If one makes the sum of the terms, on the basis of an edge of the triangle and going down in diagonal towards the line, one obtains the term located under the last term of the diagonal.
• : Example: descent of 4 terms in the column of row 3: 1 + 3 + 6 + 10 =.20 term located in bottom on the right of the last term
• : Example: Descent in diagonal of 5 terms starting from the line of row 4: 1 + 5 + 15 + 35 + 70 = 126 term located under the last
• Diagonal ascending: the sum of the terms of an ascending diagonal corresponds to the one of the terms of the Suite of Fibonacci
• : Ascending diagonals of row 1,2,3,4,5: 1,1,1 + 1 = 2, 1 + 2 = 3,1 + 3 + 1 = 5
• Each line has a center of symmetry

Formulate binomial

See also: Formula of the binomial theorem

The triangle of Pascal is often used in the binomial developments. Indeed, one finds on the same line all the coefficents intervening in the development of a power of the sum of two terms.

Example: $(X+1)\; ^2=X^2+2X+1^2\; \backslash ,$ and one notes that the coefficients of each students' rag procession are those of the third line of the triangle of Pascal (the line of row 2), i.e. 1,2,1.

Généralisation: $(X+Y)\; ^n=a\_0X^nY^0+a\_1X^\; \{n-1\}\; Y+a\_2X^\; \{N2\}\; Y^2+\; \backslash \; ldots+a\_nY^nX^0\; \backslash ,$, where the coefficients are those which are on N +1e line of the triangle of Pascal (line of row N).

Thus knowing the formula of summation $(a+b)^n\; =\; \backslash \; sum\_\; \{i=0\}\; ^n\; \backslash ,\; \{N\; \backslash \; choose\; I\}\; a^\; \{nor\}\; b^\; \{I\}$, several properties appears simply.

Let us pose has = B = 1, one has $2^n\; then\; =\; \backslash \; sum\_\; \{i=0\}\; ^n\; \backslash ,\; \{N\; \backslash \; choose\; I\}\; \backslash ,\; \{\}$.

Let us pose has = 1 and B = -1, one has $0\; then\; =\; \backslash \; sum\_\; \{i=0\}\; ^n\; \backslash ,\; \{N\; \backslash \; choose\; I\}\; \backslash ,\; (-\; 1)\; ^i$.

Knowing these two equalities, of which one is an alternated sum, it comes that the sum of the terms of order 0,2,4,… in a line is $2^\; \{n-1\}$ and is equal to the sum of the terms of order 1,3,5,….

Property related to the enumeration

The number located in the column p (while counting from 0 columns) and line N (while counting from 0 lines) indicates the number of possible Combinaison S of p elements in a unit to N elements.

In line N and the column p, there is $\{N\; \backslash \; choose\; p\}\; =\; \backslash \; frac\; \{N!\}\{p!\; (Np)!\}$.

• In line N and the column p, one reads the number of times where one can hope to obtain p piles and Np faces at the time of 2ⁿ throws of a balanced part

• By multiplying a term by the row of his column and by dividing it by the row of his line, one obtains the term located in higher notch on the left
• : example the term in line 6 and column 4 is 15 (it is pointed out that the lines and the columns are numbered while starting to 0); However 15 × 4/6=10 situédans the box right at side in top on the left.
• By multiplying the term of line N and column p par$\backslash \; frac\; \{Np\}\; \{p+1\}$, one obtains his neighbor on the line
• All the terms of the line of row N (except the first and the last) is a multiple of N if and only if N is a prime number

Numbers of Catalan

See also: Number of Catalan

All the lines of even row (2n) have a central term, by dividing this term by n+1 or by removing his neighbor to him, one obtains a number of Catalan.

Triangle of Sierpinski

See also: Triangle of Sierpiński

By graying the boxes where appears an odd number and bleaching the boxes where appears an even number, one obtains an image similar to the triangle of Sierpinski. It is the same if one blackens all the boxes which are not adequate with 0 modulo p.

Illustrated numbers

See also: Number illustrated

The numbers located on the third downward diagonal correspond to the triangular numbers those of the fourth diagonal to the tetrahedral numbers, those of the fifth diagonal to the pentatopic numbers and those of the nth diagonal to the N-topics numbers.

Trigonometrical formulas

The formula of the binomial applied to the Formula of Euler $cos\; (\backslash \; theta)\; +\; I\; \backslash \; sin\; (\backslash \; theta))^n\; =\; \backslash \; cos\; (N\; \backslash \; theta)\; +i\; \backslash \; sin\; (N\; \backslash \; theta)$ makes it possible to develop cos (nθ) and sin (nθ) The coefficients located on the line of row N make it possible to write tan (nθ) according to T = tan (θ)

Example: on line 4 one reads 1 - 4 - 6 - 4 - 1 and $\backslash \; tan\; (4\; \backslash \; theta)\; =\; \backslash \; frac\; \{4t\; -4t^3\}\; \{1-6t^2\; +\; t^4\}$

general Formule: $\backslash \; tan\; (N\; \backslash \; theta)\; =\; \backslash \; frac\; \{\backslash \; sum\_\; \{k=0\}\; ^\; \{\}\; \{(-\; 1)\; ^k\; \{N\; \backslash \; choose\; 2k+1\}\; t^\; \{2k+1\}\}\}\; \{\backslash \; sum\_\; \{k=0\}\; ^\; \{\}\; (-\; 1)\; ^k\; \{N\; \backslash \; choose\; 2k\}\; t^\; \{2k\}\}$.

The coefficients located on an ascending diagonal make it possible to express sin (N θ) like product of sin (θ) by a polynomial in cos (θ);

Example: on the ascending diagonal of row 5, one reads 1 - 3 - 1 and $\backslash \; sin\; (5\; \backslash \; theta)\; =\; \backslash \; sin\; (\backslash \; theta)\; \backslash \; left\; ((2\; \backslash \; cos\; (\backslash \; theta)\; ^4\; -\; 3\; (2\; \backslash \; cos\; (\backslash \; theta))^2\; +\; 1)\; \backslash \; right)$

Generalization: if the terms of the ascending diagonal of row N are $a\_\; \{N,\; 0\},\; a\_\; \{N,\; 1\},\; \backslash \; cdots\; a\_\; \{N,\}$, one has $\backslash \; sin\; (N\; \backslash \; theta)\; =\; \backslash \; sin\; (\backslash \; theta)\; \backslash \; left\; (\backslash \; sum\_\; \{k=0\}\; ^\; \{\}\; (-\; 1)\; ^k\; a\_\; \{N,\; K\}\; \backslash \; left\; (2\; \backslash \; cos\; (\backslash \; theta)\; \backslash \; right)\; ^\; \{n-1-2k\}\; \backslash \; right)$

Consequently, the coefficients located on the ascending diagonal of row N make it possible to determine a polynomial of degree of which the roots are the values $\backslash \; left\; (2\; \backslash \; cos\; \backslash \; left\; (\backslash \; frac\; \{K\; \backslash \; pi\}\; \{N\}\; \backslash \; right)\; \backslash \; right)\; ^2$ for K varying of 1 with : Example: on the diagonal of row 7, one reads 1 - 5 - 6 - 1, one thus knows $(2\; \backslash \; cos\; (\backslash \; frac\; \{K\; \backslash \; pi\}\; \{7\})\; ^2$ are roots of $P\; (X)\; =\; x^3\; -\; 5x^2\; +\; 6x\; -\; 1$

Généralisation: $P\; (X)\; =\; \backslash \; sum\_\; \{k=0\}\; ^\; \{(-\; 1)\; ^ka\_\; \{N,\; K\}\; x^\; \{[(n-1)\; /2\; -\; K\}$ has as roots $\backslash \; left\; (2\; \backslash \; cos\; \backslash \; left\; (\backslash \; frac\; \{K\; \backslash \; pi\}\; \{N\}\; \backslash \; right)\; \backslash \; right)\; ^2$

Generalizations

The formula of the Binôme generalized is an important generalization of the triangle of Pascal, because it makes it possible to handle complex numbers in the base, just like to use complex exhibitors.

The triangle of Pascal spreads easily with higher dimensions. The three-dimensional version is called the Pyramide of Pascal.

Appendices

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