What properties do the numbers of Pascal's triangle have? Math I like

History of the triangle. The first mention of a triangular sequence of binomial coefficients called meru-prastaara occurs in a commentary by the 10th century Indian mathematician Halayudha on the works of another mathematician, Pingala. The triangle was also explored by Omar Khayyam around 1100, which is why in Iran this pattern is called the Khayyam triangle. In 1303, the book “The Jasper Mirror” was published four elements" by Chinese mathematician Zhu Shijie, in which Pascal's triangle was depicted in one of the illustrations; It is believed to have been invented by another Chinese mathematician, Yang Hui (which is why the Chinese call it Yang Hui's triangle). On title page An arithmetic textbook written in 1529 by Peter Apian, an astronomer at Ingoltstadt University, also depicts Pascal's triangle. And in 1653 (in other sources in 1655) Blaise Pascal’s book “Treatise on the Arithmetic Triangle” was published.

Properties Pascal's triangle. If you outline Pascal's triangle, you get isosceles triangle. In this triangle, there are ones at the top and on the sides. Each number is equal to the sum of the two numbers above it. The triangle can be continued indefinitely. The lines of the triangle are symmetrical with respect to vertical axis. Has application in probability theory has entertaining properties.

Properties of Pascal's triangle. The numbers of a triangle are symmetrical (equal) about the vertical axis. first and last number are equal to 1. the second and penultimate numbers are equal to n. the third number is equal to the triangular number, which is also equal to the sum of the numbers of the previous lines. the fourth number is tetrahedral. The sum of the numbers of the ascending diagonal starting from the first element of the (n-1)th row is nth number Fibonacci. If subtracted from central number in a line with an even number, an adjacent number from the same line, then you get the Catalan number. Sum nth numbers The rows of Pascal's triangle are equal to 2n. Prime factors numbers of Pascal's triangle form symmetrical self-similar structures. If in Pascal's triangle everything odd numbers Color the even ones black and the even ones white, then a Sierpinski triangle is formed. All numbers in the nth row, except ones, are divisible by the number n if and only if n is prime number. If in a row with an odd number we add all the numbers from serial numbers of the form 3n, 3n+1, 3n+2, then the first two sums will be equal, and the third will be 1 less. Each number in the triangle is equal to the number of ways to get to it from the vertex, moving either right-down or left-down.

The famous American scientist Martin Gardner said: “Pascal’s triangle is so simple that even a ten-year-old child can write it down. At the same time, it conceals inexhaustible treasures and binds together various aspects mathematicians who at first glance have nothing in common with each other. Such unusual properties allow us to consider Pascal’s triangle one of the most elegant schemes in all of mathematics.”

Consider the following expressions with powers (a + b) n, where a + b is any binomial and n is an integer.

Each expression is a polynomial. You can notice features in all expressions.

1. In each expression there is one more term than the exponent n.

2. In each term the sum of the powers is equal to n, i.e. the power to which a binomial is raised.

3. The powers start from the binomial power n and decrease towards 0. The last term does not have a factor a. The first term does not have a factor b, i.e. degrees b start at 0 and increase to n.

4. The coefficients start at 1 and increase by certain values until “half way”, and then decrease by the same values back to 1.

Let's take a closer look at the coefficients. Let's say we want to find the value of (a + b) 6 . According to the feature we just noticed, there should be 7 members here
a 6 + c 1 a 5 b + c 2 a 4 b 2 + c 3 a 3 b 3 + c 4 a 2 b 4 + c 5 ab 5 + b 6 .
But how can we determine the value of each coefficient, c i ? We can do this in two ways. The first method involves writing the coefficients in a triangle, as shown below. This is known as Pascal's triangle :

There are many features in the triangle. Find as many as you can.
You may have found a way to write the next string of numbers using the numbers on the line above. The units are always located on the sides. Each remaining number is the sum of the two numbers above that number. Let's try to find the value of the expression (a + b) 6 by adding the following line, using the features we found:

We see that in the last line

first and last numbers 1 ;
the second number is 1 + 5, or 6 ;
the third number is 5 + 10, or 15 ;
the fourth number is 10 + 10, or 20 ;
the fifth number is 10 + 5, or 15 ; And
the sixth number is 5 + 1, or 6 .

So the expression (a + b) 6 will be equal to
(a + b) 6 = 1 a 6 + 6 a 5 b + 15 a 4 b 2 + 20 a 3 b 3 + 15 a 2 b 4 + 6 ab 5 + 1 b 6.

To raise to the power (a + b) 8, we add two lines to Pascal's triangle:

Then
(a + b) 8 = a 8 + 8a 7 b + 28a 6 b 2 + 56a 5 b 3 + 70a 4 b 4 + 56a 3 b 5 + 28a 2 b 6 + 8ab 7 + b 8 .

We can summarize our results as follows.

Newton's binomial using Pascal's triangle

For any binomial a+ b and any natural number n,
(a + b) n = c 0 a n b 0 + c 1 a n-1 b 1 + c 2 a n-2 b 2 + .... + c n-1 a 1 b n-1 + c n a 0 b n ,
where the numbers c 0 , c 1 , c 2 ,...., c n-1 , c n are taken from the (n + 1) series of Pascal’s triangle.

Example 1 Raise to a power: (u - v) 5 .

Solution We have (a + b) n , where a = u, b = -v, and n = 5. We use the 6th row of Pascal's triangle:
1 5 10 10 5 1
Then we have
(u - v) 5 = 5 = 1 (u)5+ 5 (u) 4 (-v) 1 + 10 (u) 3 (-v) 2 + 10 (u) 2 (-v) 3 + 5 (u)(-v) 4 + 1 (-v) 5 = u 5 - 5u 4 v + 10u 3 v 2 - 10u 2 v 3 + 5uv 4 - v 5 .
Note that the signs of the terms fluctuate between + and -. When the degree -v is an odd number, the sign is -.

Example 2 Raise to a power: (2t + 3/t) 4 .

Solution We have (a + b)n, where a = 2t, b = 3/t, and n = 4. We use the 5th row of Pascal's triangle:
1 4 6 4 1
Then we have

Binomial expansion using factorial values

Let's say we want to find the value of (a + b) 11. The disadvantage to using Pascal's triangle is that we have to calculate all the previous rows of the triangle to get required row. Next method allows you to avoid this. It also allows you to find a specific row - say the 8th row - without having to evaluate all the other rows. This method is useful in calculations, statistics and it uses binomial coefficient notation .
We can formulate Newton's binomial as follows.

Newton's binomial using factorial notation

For any binomial (a + b) and any natural number n,
.

Newton's binomial can be proven by the method mathematical induction. It shows why it's called binomial coefficient .

Example 3 Raise to a power: (x 2 - 2y) 5 .

Solution We have (a + b) n , where a = x 2 , b = -2y, and n = 5. Then, using Newton's binomial, we have

Finally, (x 2 - 2y) 5 = x 10 - 10x 8 y + 40x 6 y 2 - 80x 4 y 3 + 80x 2 y 4 - 35y 5 .

Example 4 Raise to a power: (2/x + 3√x) 4.

Solution We have (a + b)n, where a = 2/x, b = 3√x, and n = 4. Then, using Newton's binomial, we get

Finally (2/x + 3√x ) 4 = 16/x 4 + 96/x 5/2 + 216/x + 216x 1/2 + 81x 2 .

Finding a specific member

Let's assume that we want to determine one or another term from an expression. The method we have developed will allow us to find this term without calculating all the rows of Pascal's triangle or all the previous coefficients.

Note that in Newton's binomial gives us the 1st term, gives us the 2nd term, gives us the 3rd term and so on. This can be summarized as follows.

Finding the (k + 1) term

(k + 1) term of the expression (a + b) n is .

Example 5 Find the 5th term in the expression (2x - 5y) 6 .

Solution First, note that 5 = 4 + 1. Then k = 4, a = 2x, b = -5y, and n = 6. Then the 5th term of the expression will be

Example 6 Find the 8th term in the expression (3x - 2) 10.

Solution First, we note that 8 = 7 + 1. Then k = 7, a = 3x, b = -2 and n = 10. Then the 8th term of the expression will be

Total number of subsets

Suppose a set has n objects. The number of subsets containing k elements is . The total number of subsets of a set is the number of subsets with 0 elements, as well as the number of subsets with 1 element, and also the number of subsets with 2 elements, and so on. The total number of subsets of a set with n elements is
.
Now let's look at raising to the power (1 + 1) n:

.
So. the total number of subsets is (1 + 1) n, or 2 n. We have proven the following.

Total number of subsets

The total number of subsets of a set with n elements is 2n.

Example 7 How many subsets does the set (A, B, C, D, E) have?

Solution The set has 5 elements, then the number of subsets is 2 5, or 32.

Example 8 Wendy's restaurant chain offers the following hamburger toppings:
{ketchup, mustard, mayonnaise, tomatoes, lettuce, onion, mushrooms, olives, cheese}.
How many different types What burgers can Wendy offer, excluding the size of the burgers or the number of burgers?

Solution The toppings on each hamburger are members of a subset of the set of all possible toppings, and the empty set is just a hamburger. The total number of possible hamburgers will be equal to

. Thus, Wendy's can offer 512 different hamburgers.

Published in Hard"n"Soft magazine No. 10 2003

The amazing triangle of the great Frenchman

I remember well one professor who had
vision and thought he was going crazy.
He came to me in a state of complete panic.
In response, I simply took a book from the shelf written by
about four hundred years ago, and showed the patient
wood engraving depicting exactly
what he imagined.
Carl Gustav Jung. Man and his symbols.

When I read Pascal, it seems to me
that I am reading myself.
Stendhal

Derogatory wording" irreplaceable people no,” so beloved by incompetent managers, might be suitable if we were talking about digging a trench or cleaning up garbage. Any type of activity related to creativity, on the contrary, will show the irreplaceability and uniqueness of each person. And when we're talking about about geniuses, then we should all thank fate for the opportunity to enjoy the fruits of their activity, for the light emanating from them, illuminating the paths of human development. On the website of the magazine "Knowledge is Power" there is a vote on who you consider the most significant scientist of the past 2000 years. (http://www.znanie-sila.ru/vote/?id=2 - look, by the way, it’s interesting to compare your preferences with the choice of the majority.) And, naturally, among the most popular scientists we rightfully see the name of Blaise Pascal (1623 -1662).

Pascal died when he was 39 years old, but despite such short life, he went down in history as outstanding mathematician, physicist, philosopher and writer. A unit of pressure (pascal) and an extremely widespread programming language are named after him by his grateful descendants. Turbo Pascal 5.5 for DOS was especially popular, now Borland Pascal 7.0 and its further development in Delphi. Pascal's works span the most different areas. He is one of the creators mathematical analysis, projective geometry, probability theory, hydrostatics (Pascal’s law is widely known, according to which a change in pressure in a fluid at rest is transmitted to its other points without changes), the creator of a mechanical calculating device - the “Pascal wheel” - as contemporaries said. Pascal demonstrated that air has elasticity, proved that it has weight, and discovered that barometer readings depend on humidity and air temperature and therefore can be used to predict the weather.

Some of practical achievements Pascal was awarded highest distinction- today few people know the name of their author. For example, now very few people will say that the most ordinary car is the invention of Blaise Pascal. He also came up with the idea of omnibuses - multi-seat horse-drawn carriages with fixed routes - the first type of regular public urban transport. Already at the age of sixteen, Pascal formulated a theorem about a hexagon inscribed in conic section(Pascal's theorem). (It is known that he later obtained about 400 corollaries from his theorem.) A few years later, Blaise Pascal created a mechanical computing device - a summing machine that made it possible to add numbers into decimal system Reckoning In this machine, numbers were set by corresponding turns of disks (wheels) with digital divisions, and the result of the operation could be read in windows - one for each digit.

Blaise Pascal and another great Frenchman, Pierre Fermat, became the founders of the theory of probability, and the year of its birth is often called 1654, when Pascal and Fermat independently gave correct explanation the so-called rate division paradox. Two players play a "harmless" game (that is, both have the same chance of winning), agreeing that the first to win six games will receive the entire prize. Let's assume that the game stopped before one of them won a prize (for example, the first player won five games, and the second player won three). How to divide the prize fairly? Although, generally speaking, this problem is not a paradox, the unsuccessful attempts of some prominent scientists to solve it, as well as incorrect answers, created the legend of the paradox. Thus, according to one decision, the prize should have been divided in a ratio of 5: 3, i.e. in proportion to the games won, according to another - in the ratio 2: 1 (here the reasoning was apparently carried out as follows: since the first player won two more games, which is a third of the six games needed to win, he should receive one third from the prize, and the remainder must be divided in half).

Meanwhile, you need to divide in a ratio of 7:1. Both Pascal and Fermat treated the bet division paradox as a probability problem, establishing that a fair division was proportional to the first player's chances of winning the prize. Suppose the first player has only one game left to win, and the second needs to win three more games to win, and the players continue the game and play all three games, even if some of them turn out to be unnecessary to determine the winner. For such a continuation, all 2 3 = 8 possible outcomes will be equally probable. Since the second player receives a prize in only one outcome (if he wins all three games), and in other cases the first player wins, the ratio is 7: 1. (Pascal and Fermat also found general solution for the case when one player needs to win n more games to receive a prize, and the other needs to win m games.)

But perhaps Blaise Pascal's most famous mathematical work is his treatise on the "arithmetic triangle" formed by binomial coefficients (Pascal's triangle), which has applications in probability theory and has surprising and entertaining properties. We will consider this magic triangle; those who want to deepen their knowledge about the brilliant scientist will find a list of literature about him on http://inf.1september.ru/2002/1/france.htm, and on the “Submarine” http://schools. techno.ru/sch444/MUSEUM/PRES/PL-4-98.htm an intriguing story about Pascal, his father, sister and Cardinal Richelieu himself.

The triangle will be drunk
You give it with a bang!
Even if he were a parallelepiped,
If he were a cube, he'd be a louse
V.Vysotsky

In fact, Pascal's triangle was known long before 1653, the date of publication of the Treatise on the Arithmetic Triangle. Thus, this triangle is reproduced on the title page of an arithmetic textbook written in early XVI Peter Apian, an astronomer at Ingoltstadt University. A triangle is also depicted in an illustration in a book by a Chinese mathematician published in 1303. Omar Khayyam, who was not only a philosopher and poet, but also a mathematician, knew about the existence of the triangle around 1100, in turn borrowing it from earlier Chinese or Indian sources.

Martin Gardner writes in the book “Mathematical Novels” (M., Mir, 1974): “Pascal’s triangle is so simple that even a ten-year-old child can write it down. At the same time, it conceals inexhaustible treasures and connects together various aspects of mathematics that have no at first glance there is nothing in common with each other. Such unusual properties allow us to consider Pascal’s triangle one of the most elegant schemes in all of mathematics."

Let's assume that you enter the city as shown in the diagram with the blue arrow, and you can only move forward, or rather, constantly choosing, forward to the left, or forward to the right. Nodes that can only be reached in one way are marked with green smiley faces; a point that can be reached in two ways is shown with a red smiley face, and with three, respectively, in pink. This is one of the options for constructing a triangle, proposed by Hugo Steinhaus in his classic "Mathematical Kaleidoscope".

And the structure of Pascal’s triangle is explained even more simply by the following words: each number is equal to the sum of the two numbers located above it. Everything is elementary, but there are so many miracles hidden in it.

The vertex of the triangle is 1. The triangle can be continued indefinitely. It is symmetrical about a vertical axis passing through its apex. Along the diagonals (as far as a triangle can have diagonals, but let’s not quibble, such terminology is found in publications), parallel to the sides triangle (marked with green lines in the figure) triangular numbers and their generalizations to the case of spaces of all dimensions are constructed.

Triangular numbers in the most common and familiar form show how many touching circles can be arranged in the form of a triangle - how classic example initial arrangement of balls in billiards. You can attach two more to one coin - for a total of three - to two you can attach three more - for a total of six. Continuing to increase the rows while maintaining the shape of the triangle, we get row 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66..., which is what the second shows green line. This wonderful series, each member of which is equal to the sum of the natural series of numbers (55=1+2+3+4+5+6+7+8+9+10), also contains many familiar ones that are well known to mathematics lovers: 6 and 28 - perfect numbers, 36 - square number, 8 and 21 are Fibonacci numbers.

The next green line will show us tetrahedral numbers - we can put one ball on three - a total of four, under three we can put six (strain yourself and imagine!) - a total of ten, and so on. You can read more about triangular numbers in Hard"n"Soft No. 4 2002 in the article "Cannibal Rabbits, Quatrains and the Reserve of Sequences" also available on Watermelon.

And the next green line (1, 5, 15, 35,...) will demonstrate an attempt to lay out a hypertetrahedron in four-dimensional space- one ball touches four, and they, in turn, touch ten... In our world this is impossible, only in a four-dimensional, virtual one. And even more so, the five-dimensional tetrahedron, which is evidenced by the next green line, can exist only in the reasoning of topologists.

But what does the top green line, on which the numbers of the natural series are located, tell us? These are also triangular numbers, but one-dimensional, showing how many balls can be laid out along the line - as many as there are, lay out as many. If we go to the end, then the topmost row of ones are also triangular numbers in zero-dimensional space - no matter how many balls we take, we will not be able to place more than one, because there is simply nowhere - there is no length, no width, no height.

Even a quick glance at Pascal's triangle is enough to note the following curious facts: 10 nuclei can be folded both in the form of a tetrahedron and in the form of a flat triangle. And 56 hypernuclei forming a tetrahedron in five-dimensional space can be laid out in the usual familiar three-dimensional tetrahedron, however, if we tried to lay out a triangle out of 56 nuclei, then one core would remain extra.

How can we draw Pascal's triangle to play with it? It is best to use the idea that we considered when programming hexagonal life in Hard"n"Soft No. 5 2002 (on Arbuz), namely, an ordinary two-dimensional array is taken, but when displayed on the screen, the rows are shifted after one - even rows to the right by a quarter step, odd ones to the left by a quarter step, and then the rows are shifted by half a step, which gives us a hexagonal field structure with a rectangular array. And the two-dimensionality of the array makes it very easy to work with it, specifying actions on the cell in a loop in rows and rows.

After fiddling around for a couple of minutes, you will be rewarded with a triangle appearing on the screen and, therefore, you are ready for the upcoming unusual experiments. (You shouldn’t specify too many rows, since from 13-14 rows four and five digit numbers, they merge with those standing next to them and the picture becomes blurred. You can, of course, increase the cell radius and reduce the font, but still, the numbers in the middle of the triangle grow quickly and will merge, albeit a couple of rows lower.)

But first a couple more interesting properties Pascal's triangle. To find the sum of numbers on any diagonal from the beginning to the place of interest to us, just look at the number located below and to the left of the last term. (on the left for the right diagonal, for the left diagonal it will be on the right, and in general - closer to the middle of the triangle). Let, for example, we want to calculate the sum of the numbers in the natural series from 1 to 9. “Going down” diagonally to the number 9, we will see the number 45 to the lower left of it. This is what gives the required sum. What is the sum of the first eight triangular numbers? We find the eighth number on the second diagonal and move down and to the left. Answer: 120. But, by the way, 120 is a tetrahedral number. Therefore, by taking all the balls that make up the first 8 triangles, we could form a tetrahedron. Try with cherries or apples same size, just don’t try to go into the fourth dimension with them, they may disappear.

The sums of numbers along the not so steeply falling diagonals (marked with red lines in the figure) form the Fibonacci sequence, well known to regular readers. See, for example, the above-mentioned article “Cannibal Rabbits, Quatrains...” or numerous materials on Watermelon.

But in previous publications we did not talk about the fact that Fibonacci numbers are often found in combinatorial problems Oh. Consider a row of n chairs. In how many ways can men and women be seated on them so that no two women sit next to each other? When n=1, 2, 3, 4, ... the number of ways is respectively 2, 3, 5, 8, ..., that is, it coincides with the Fibonacci numbers. Pascal apparently did not know that the Fibonacci numbers were hidden in his triangle. This circumstance was discovered only in the 19th century. The numbers on the horizontal lines of Pascal's triangle are binomial coefficients, that is, expansion coefficients (x+y) n in powers of x and y. For example, (x+y) 2 =x 2 +2xy+y 2 and (x+y) 3 =x 3 +3x 2 y+3xy2+y 3. The expansion coefficients 1, 2, 2 are in the second row, and 1, 3, 3, 1 are in the third row of the triangle. To find the expansion coefficients (x+y) n, just look at the nth row of the triangle. This is exactly what fundamental property Pascal's triangle connects it with combinatorics and probability theory, turning it into a convenient means of performing calculations.

Suppose (an example from Martin Gardner) that a certain sheikh, following the laws of hospitality, decides to give you three of his seven wives. How many different choices can you make among the beautiful harem dwellers? To answer this exciting question you just need to find the number at the intersection of diagonal 3 and line 7: it turns out to be equal to 35. If, overcome with joyful excitement, you confuse the diagonal and line numbers and look for the number at the intersection of diagonal 7 with line 3, you will find that they do not intersect. That is, the method itself does not allow you to make mistakes!

IN general case, a number indicating how many ways n elements can be selected from a set containing r various elements, stands at the intersection of the n-th diagonal and the r-th line. And once again, for those who at least understood something. The number of possible combinations of n elements by m is determined by the formula

Where n!=1*2*3*4*....*n is the so-called factorial of the number n. And the same three wives out of seven can be chosen in so many ways: C 3 7 =7!/3!/4!=1*2*3*4*5*6*7/1*2*3/1*2*3 *4=5040/6/24=35, which is what we got earlier. And the values of the binomial coefficients are determined by the formula and, they are, as we found out, the rows of Pascal’s triangle, incomprehensibly connecting this triangle with combinatorics and the expansion of binomials in powers.

By the way, from the combination formula it follows that the number of options for choosing three out of seven is equal to the number of options for choosing four out of seven, or, the number of options for filling out Sportloto cards 5 out of 36 is equal to the number of choosing 31 out of 36, think about this pleasant subject.

The connection between combinatorics and probability theory becomes clear if we consider the eight possible outcomes of tossing three coins: GGG, GGR, GRG, RGG, RGR, RRG, RRR. It is not difficult to see that three coats of arms appear in only one case, two coats of arms in three cases, one coat of arms in three cases, and no coats of arms in one case. The numbers of favorable tests for receiving 3, 2, 1 and 0 coats of arms are 1, 3, 3, 1. These are the numbers that appear in the third row of Pascal's triangle. Now suppose that we want to know the probability of getting exactly 5 coats of arms when throwing 10 coins at the same time. First of all, you need to count how many there are in various ways, allowing you to choose 5 coins out of 10. We get the answer by finding the number at the intersection of the 5th diagonal and the 10th line. It is equal to 252. By adding up all the numbers in the 10th line, we find the number of possible outcomes; calculations can be greatly reduced if we use the following property of binomial coefficients: the sum of the binomial coefficients (x+y) n, and it is they that stand in n The th row of Pascal's triangle is equal to 2 n. Really, sum of numbers, standing in any row of the triangle, is twice the sum of the numbers in the previous line, since when constructing each line, the numbers in the previous one are taken down twice. The sum of the numbers in the first (topmost) row is equal to 1. Therefore, the sums of the numbers in the rows of Pascal's triangle form a geometric progression with the first term equal to 1 and the denominator 2: 1, 2, 4, 8, .... The tenth power of 2 is 1024. Therefore, the probability of getting five heads when throwing 10 coins is 252/1024= 63/256. Those wishing to learn more about the connection between Pascal's triangle and combinatorics can visit the page http://combinatorica.narod.ru/third.html.

Pascal's triangle is two-dimensional and lies in a plane. Soap appears involuntarily - but is it possible to extend its patterns to a three-dimensional (and four-...) analogue? It turns out it is possible! The article by O. V. Kuzmin (http://www.pereplet.ru/obrazovanie/stsoros/1006.html) considers a three-dimensional analogue of a triangle - Pascal’s pyramid, its connection with trinomial coefficients and gives process examples, which such a model can reflect.

Now, finally, let's move on to the most interesting part for us. amazing property Pascal's triangle. Let's replace each number in Pascal's triangle with a dot. Moreover, we will display the odd points in a contrasting color, and the even ones in a transparent or background color. The result will be unexpectedly surprising: Pascal's triangle will break into smaller triangles, forming an elegant pattern. These patterns are fraught with many surprises. As we move away from the vertex, we will encounter triangles of ever increasing sizes, not containing a single bold point, that is, “composed” of only even numbers. At the vertex of Pascal’s triangle there is a triangle “hidden” consisting of one single point, then there are triangles containing 6, 28, 120, 496, ... points. Three of these numbers - 6, 28 and 496 - are known to be perfect because each of them is equal to the sum of all its divisors other than the number itself. For example, 6=1+2+3. It is unknown whether there are an infinite number of perfect numbers, or whether there is even one odd perfect number. You can read more about perfect numbers on Watermelon.

Let us present a program that implements the coloring of a triangle in accordance with the parity of each number of the triangle. Instead of the value of the number, a circle is drawn in its place, filled with black for odd values and white for even values.

The parity of a number can be easily determined by comparing the remainder of division by two with zero; for an even number the remainder is zero, for an odd number the remainder is one. And to determine the remainder, you can use the Mod function, available in almost all programming languages. If you are too lazy to program, but definitely want to see this miracle, then go to http://www.informika.ru/text/inftech/edu/edujava/mathematics/Pascal/Pascal.html and you will find there an applet that draws Pascal’s triangle with dots taking into account parity.

There is also a link to the source code in Java, you can understand and improve it at your own discretion. Math lovers will immediately be struck by the “fractality” of the resulting object, or rather, we see nothing more than the “Sierpinski Triangle”, an analogue of the famous “Sierpinski Carpet”. These models are especially popular, along with the “Koch Snowflake” and the Mandelbrot and Julie Steel sets in recent years due to the craze for fractals and synergetics. Let us explain briefly for beginners.

From the master of popular mathematics Martin Gardner we find that back in 1905 at the annual Mathematical Olympiad in Hungary, the problem was proposed: “A square is divided into 9 parts (as for the game tic-tac-toe) and the central square is removed. Then each of the remaining 8 squares is divided into 9 parts, the central square is removed and the procedure is repeated many times. Find the limit to which the area tends the resulting figure." So - the resulting figure is the Sierpinski carpet - the square is so full of holes that it is already closer to the line. The triangle we saw can be obtained in the same way - initially the midpoints of the sides of the triangle are connected and the resulting triangle is removed.

At the second stage, the same operation is carried out with the three remaining triangles, then with the remaining nine, and so on. Can you find the limit to which the remaining area tends? And how to explain the coincidence of the two models?

The authors of the page http://chaos.h1.ru/ChaosAndFractals/1/ propose to immediately build Pascal’s triangle by filling it not with numbers, but with zeros or ones according to the rule: the sum of two zeros or two ones gives zero (that is, the sum of two even or two odd numbers is always even), and the sum of zero and one gives one (like the sum of an even number with an odd one). This technique will allow us to build an arbitrarily large triangle, and when filling it with “real” numbers, we may encounter a limitation on machine representation numbers, and with the fact that the Mod function is on limit of number declared as Double starts to fail. The authors of the mentioned page also propose to organize the triangle as a two-dimensional array (which is what we did) and use its field to model Cellular Automata, which is what we did in the article about the game Life (on Watermelon), although without limiting the field to a triangle.

We move on - we try to check not the parity, but the remainder of division by other numbers, and each time we are surprised by the appearance of the triangle. After playing for a while, we will notice that when you set the number by which we are checking to be a simple number, you get beautiful patterns with a pronounced pattern (try setting 3, 5, 7, 11, 13, 17...), and when dividing by composite number the ornament crumbles, maintaining, however, symmetry and regularity in the alternation of patterns. Moreover, the more divisors the number being tested has (for example, 12 is divided by 2, 3, 4 and 6), the more “blurred” the pattern turns out.

Consider a triangle constructed “relatively” to the number 7, that is, numbers that are not divisible by 7 without a remainder are drawn in black, those that are divisible are drawn in white, and try to see the patterns.

For those wishing to delve deeper into the connections between combinatorics, probability theory and Pascal's triangle, we recommend Gregory J. Chaitin's article "Randomness in Arithmetic" from the journal IN THE WORLD OF SCIENCE. (Scientific American. Edition in Russian). No. 9 1988, located at http://grokhovs2.chat.ru/arith/arith.html, but for now we’ll do something new - let’s try to color Pascal’s triangle. To do this, we will assign three variables (r,g,b) responsible, respectively, for the red, green and blue components of the cell coloring and tie their value (the maximum can be equal to 255) to the divisibility check by different numbers. In the above program listing, the red color depends, as before, on the parity of the number, the green color depends on its divisibility by 9, and the blue color depends on its divisibility by 11. Numerous variants of experiments are marked with apostrophes as comments, you can “revive them” or come up with your own " control numbers" and their color shades.

Dim a(100, 100) As Double Dim radius As Byte, i As Byte, kol As Byte Dim sdvig As Integer, X As Integer, Y As Integer, X1 As Integer, Y1 As Integer Private Sub Form_Load() For Y = 1 To kol For X = 1 To kol a(X, Y) = 0 Next X Next Y radius = 5 " cell radius in pixels kol = 20 " Number of rows a(Int(kol / 2), 0) = 1 " first unit , from which the triangle grows DrawWidth = 1 "Line thickness For Y = 0 To kol For X = 1 To kol sdvig = radius / 2 * (-1) ^ Y " Shift each row to the left, then to the right If Y > 0 Then If sdvig > 0 Then a(X, Y) = a(X + 1, Y - 1) + a(X - 0, Y - 1) Else a(X, Y) = a(X + 0, Y - 1 ) + a(X - 1, Y - 1) End If End If X1 = 60 + X * radius * 2 + sdvig Y1 = 10 + Y * radius * 1.7 FillStyle = 0 r = 0: g = 0: b = 0 If a(X, Y) > 0 Then If (a(X, Y) - Int(a(X, Y) / 2) * 2) = 0 Then r = 250 "If (a(X, Y) / 4 ) - Int(a(X, Y) / 4) = 0 Then r = 120 "If (a(X, Y) / 8) - Int(a(X, Y) / 8) = 0 Then r = 180 " If (a(X, Y) / 16) - Int(a(X, Y) / 16) = 0 Then r = 250 "If (a(X, Y) / 3) - Int(a(X, Y) / 3) = 0 Then g = 60 If (a(X, Y) / 9) - Int(a(X, Y) / 9) = 0 Then g = 250 "If (a(X, Y) / 7) - Int(a(X, Y) / 7) = 0 Then g = 180 "If (a(X, Y) / 5) - Int(a(X, Y) / 5) = 0 Then g = 250 If ( a(X, Y) / 11) - Int(a(X, Y) / 11) = 0 Then b = 250 "If (a(X, Y) / 13) - Int(a(X, Y) / 13 ) = 0 Then b = 120 "If (a(X, Y) / 17) - Int(a(X, Y) / 17) = 0 Then b = 180 "If (a(X, Y) / 19) - Int(a(X, Y) / 19) = 0 Then b = 250 ForeColor = RGB(r, g, b) FillColor = RGB(r, g, b) " Fill color Circle (X1, Y1), radius, RGB (90, 90, 90) End If Next X Next Y " Exit the program Private Sub Exit_Click() End End Sub

And here is the result of the program. Isn't it beautiful? Red triangular “Sierpinski zones” are visible, which, superimposed on the green windows from nines, give yellow zones, and with blue areas from division by 11 give lilac areas. Does this beauty have applied value except for the pattern for the wallpaper, it is not yet clear, but from Pascal’s triangle, especially the colored one, one can expect any miracles, perhaps in the near future. And here's another one coloring option, performed according to the algorithm

R = a(x, y) / 3 Mod 255 g = a(x, y) / 2 Mod 255 b = a(x, y) / 4 Mod 255

Look at the picture, try to link it with the algorithm, or even better, try your own version. The article http://www.webbyawards.ru/pcworld/2001/07/130_print.htm suggests using recursion to construct Pascal's triangle. What recursion is, and how optimal it is for programming, can be found at http://arbuz.ferghana.ru/z_vetki.htm. On the pages http://hcinsu.chat.ru/algoritm/mathem/binom.html and http://dkws.narod.ru/math/tpas.html you will find programs for composing Pascal’s triangle, and on the page http:// galibin.chat.ru/Java/Pascal/index.html there is also an applet that draws it on the screen, however, you are now already fully armed, but these pages can give you new ideas.

There is more about Pascal's triangle good article host of the entertaining programming column "Computer News" A. Kolesnikov at http://www.kv.by/index2002151201.htm. We began our consideration of Pascal's triangle with movement options, and we will finish with them. There is a book on the page dedicated to puzzles Evgenia Gika "Chess and Mathematics". In the chapter devoted to the geometry of the chessboard (http://golovolomka.hobby.ru/books/gik/05.shtml) the author gives amazing examples, when knowledge of the king's route options allowed the masters to save completely losing positions. (The famous study by Reti is shown, in which the king miraculously manages to fight in two opposite sections of the board.) And the connection with our topic is that the number of options for the king’s routes to reach each square obeys the law of Pascal’s triangle! See the diagram, as they say in chess textbooks. And use it in your endgames.

And the very last question, related both to Pascal’s triangle and to chess. What is the sum of all numbers above any row? Consider these sums for yourself, starting from the top, and you will see the values 1, 3, 7, 15, 31,... You don’t need to have much imagination to see a simple pattern: the sum of all numbers for n rows is 2 n -1. And what does chess have to do with it? According to a well-known legend, the Raja promised the creator of chess any reward he asked for. When the first chess player asked to put one grain of wheat on the first square of the board, two on the second, four on the third, and so on, continuing to double it, until the 64th square, the Raja was even offended at first by the meagerness of the requested reward. When his storekeepers estimated the requested quantity, it turned out that this grain could cover the entire Earth knee-deep; this is much more than has been and will be collected in all the harvests of mankind. (By the way, you can estimate the height of the grain layer, given the volume of the grain, for example, 1 mm 3, multiply by 2 64, certainly subtract 1 and divide by the area of the earth's surface.) So - on each square of the board there would be (would be) the number of grains, equal to the sum numbers in the corresponding row of Pascal's triangle, and the sum of all the grains on the first n cells would be equal to the sum of the numbers on these n rows of this magic triangle. With this abundant fantasy we will conclude our consideration.

Variations on the theme "Pascal's Triangle"

Story

Pascal's triangle is perhaps one of the most famous and elegant number schemes in all of mathematics.

Blaise Pascal, a French mathematician and philosopher, dedicated a special “Treatise on the Arithmetic Triangle” to her.

However, this triangular table was known long before 1665 - the date of publication of the treatise.

Thus, in 1529, Pascal's triangle was reproduced on the title page of an arithmetic textbook written by the astronomer Peter Apian.

A triangle is also depicted in the illustration of the book “The Jasper Mirror of the Four Elements” by the Chinese mathematician Zhu Shijie, published in 1303.

Omar Khayyam, who was not only a philosopher and poet, but also a mathematician, knew about the existence of the triangle in 1110, in turn borrowing it from earlier Chinese or Indian sources.

Construction of Pascal's triangle

Pascal's triangle is simply an infinite number table "triangular shape", in which there are ones at the top and on the sides, each of the remaining numbers is equal to the sum of the two numbers above it on the left and on the right in the previous line. The table has symmetry about the axis passing through its top.

Properties of Pascal's triangle

String Properties

nth line

next property

Triangular numbers
Triangular, tetrahedral and other numbers are lined up along the diagonals parallel to the sides of the triangle. Triangular numbers indicate the number of balls or other objects arranged in the form of a triangle (these numbers form the following sequence: 1,3,6,10,15,21,..., in which 1 is the first triangular number, 3 is the second triangular number, 6-third, etc. up to m-ro, which shows how many terms of Pascal’s triangle are contained in the first m lines - from zero to (m-1)th).

Tetrahedral numbers
The members of the sequence 1,4, 10, 20, 36, 56,... are called pyramidal, or more precisely, tetrahedral numbers: 1 is the first tetrahedral number, 4 is the second, 10 is the third, etc. up to m-ro . These numbers show how many balls can be stacked in the form triangular pyramid(tetrahedron).

Fibonacci numbers
In 1228, the outstanding Italian mathematician Leonardo of Pisa, better known now as Fibonacci, wrote his famous “Book of Abacus.” One of the problems in this book, the rabbit breeding problem, resulted in the sequence of numbers 1,1,2,3,5,8,13,21..., in which each term, starting from the third, is the sum of the two previous terms. This sequence is called the Fibonacci series, the members of the Fibonacci series are called Fibonacci numbers. Denoting the nth Fibonacci number by

There is an interesting connection between the Fibonacci series and Pascal's triangle. For each ascending diagonal of Pascal's triangle, we form the sum of all numbers on this diagonal. We get 1 for the first diagonal, 1 for the second, 2 for the third, 3 for the fourth, and 5 for the fifth. We got nothing more than five initial Fibonacci numbers. It turns out that the sum of numbers is always nth diagonal is the nth Fibonacci number. To prove the proposition we are interested in, it is enough to show that the sum of all the numbers that make up the nth and (n+1) diagonals of Pascal’s triangle is equal to the sum of the numbers that make up its m+2nd diagonal.

Binomial coefficients
The numbers on the horizontal lines are binomial coefficients. The line numbered n consists of the coefficients of the binomial expansion (1+n)n. Let's show this using Pascal's operation. But first, let's imagine how binomial coefficients are determined.

Let's take a binomial 1 + x and begin to raise it to the powers 0, 1, 2, 3, etc., arranging the resulting polynomials in increasing powers of the letter x. We'll get

1.(1+x)0=1,
2.(1+x)1=1+x,
3. (1 +x)2=(1 +x)(1 +x)= 1 +2x+x2,
4.(1+х)3=1+Зх+Зх2+хЗ
etc.

In general, for any whole non-negative number n
(1+x)n=a0+a1x+a2x2+...+apxp,
where a0,a1,...,ap

The last relation can be rewritten in the form and from relations 1-4 we obtain

Pascal's triangle was formed, each element of which

It is this fundamental property of Pascal’s triangle that connects it not only with combinatorics and probability theory, but also with other areas of mathematics and its applications.

Solving problems using Pascal's triangle

Ancient problems about randomness
Since ancient times, various gambling. IN Ancient Greece and Rome, games of astragalus became widespread, when players threw animal bones. Also popular dice- cubes with dots marked on the faces. Gambling later spread throughout medieval Europe.

These games gave mathematicians a lot of interesting tasks, which later formed the basis of probability theory. Problems about dividing the bet were very popular. After all, as a rule, the game was played for money: players made bets, and the winner took the entire amount. However, the game was sometimes interrupted before the end, and the question arose: how to divide the money.

Many mathematicians have worked on solving this problem, but until mid-17th century centuries they never found him. In 1654 between French mathematicians Blaise Pascal, already well known to us, and Pierre Fermat began a correspondence regarding a number of combinatorial problems, including problems of dividing the bet. Both scientists, although several in different ways, came to the right decision, dividing the bet in proportion to the probability of winning the entire amount if the game continues.

It should be noted that before them, none of the mathematicians calculated the probability of events; in their correspondence, the theory of probability and combinatorics were first scientifically substantiated, and therefore Pascal and Fermat are considered the founders of the theory of probability.

Let's consider one of Fermat's problems, solved by Pascal using his numerical table.

Let player A need two games before winning the entire match, and player B need three games. How to split the bet fairly if the game is interrupted?

Pascal adds up the number of games missing to the players and takes a table row in which the number of terms is equal to the found sum, i.e. 5. Then the share of player A will be equal to the sum of the three (according to the number of games missing to player B) first terms of the fifth row, and Player B's share is the sum of the remaining two numbers. Let's write this line: 1,4,6,4, 1. Player A's share is 1+4+6=11, and B's share is -1+4=5.

Other arithmetic triangles

Let us consider triangles, the construction of which is associated with known one-parameter combinatorial numbers. The creation of such triangles is based on the principle of constructing the Pascal triangle discussed above.

Luke's triangle

Consider the constructed arithmetic triangle. This triangle is called the Lucas triangle, since the sums of the numbers on the ascending diagonals give the sequence of Lucas numbers: 1, 3, 4, 7, 11, 18, / which can be defined as

Ln=Ln-1+Ln-2, L0=2, L1=1

Each element of the triangle is determined by Pascal's rule Ln+1,k=Ln, k-1+Ln, k with initial conditions L1,0=1, L1,1=2 and L0,k=0

i.e. nth line triangle hatch can be obtained adding the nth and (n-1)th rows of Pascal's triangle.

Fibonacci Triangle

From the numbers (fm, n) satisfying the equations
fm, n=fm-1,n+fm-2,n,
fm, n=fm-1,n-1+fm-2,n-2, where c initial conditions f0,0=f1,0=f1,1=f2,1=1 the next triangle is constructed.

fm, n =fn fn-m, m Є n Є 0, where fn is the nth Fibonacci number. The constructed triangle is called the Fibonacci triangle.

Tribonacci triangle

Let's consider another triangle, the creation of which is based on the method of constructing Pascal's triangle. This is the Tribonacci triangle. It is named so because the sums of the elements on the ascending diagonals form a sequence of Tribonacci numbers: 1,1,2,4,7,13,24,44,..., which can be defined by the following recurrence relation: tn+3 = tn+2 + tn+1 + tn with initial conditions t0 = 1, t1 = 1, t2 = 2

"Iconic Triangle"

Construction of the "sign triangle"

Before us is a triangle, made up of only signs, pluses and minuses, according to the principle of the formation of Pascal’s triangle. Unlike the latter, it is located base up.

First, the first line is set, consisting of an arbitrary number of characters and their location. Each character of the next line is obtained by multiplying two higher characters.

One of our tasks is to establish at what number of characters in the first line the number of minuses and pluses will be the same. Total quantity characters in the table can be determined by the formula

where n is the number of characters in the first line.

A sequence of numbers is formed in which the number of minuses and pluses can be equal: 3, 4, 7, 8, 11, 12, 15, 16,..., each of which shows the number of characters in the first line. However, it has not been established at what arrangement of signs the number of minuses and pluses will be uniquely the same.

Our second task regarding the triangle of the product of signs is to establish the smallest number of pluses that a “sign triangle” can have.

There is an interesting sequence of characters in the first line: +, -, -, +, -, -, ... (or -, -, + ,- ,- ,+ , ...), in which the number of pluses, as before is considered to be the smallest and equal to 1/3 of total number signs, i.e. equal

It is important to note that if you gradually go around the triangle, the sequence of signs +, -, -, ... will remain.

Let us pay attention to the fact that the smallest number of pluses, equal to 1/3 of the total number of signs, can also be seen in the triangle with n = 2.