Freedom dependent variables
April 1 is known as the day about white backs and untied shoelaces. But at the university where I studied, this day is also considered the day of mathematics. Therefore, I decided in this habrastitia to collect some of my amusing puzzles and stories connected in some sense with mathematics.
Imagine that the earth is a ball and we are traveling with a friend from Omsk to Moscow along an arc of a circle, then my friend, who drives on the lower shelf, will travel a path equal to the arc length B 1 B 2 , and, I am a path equal to the arc length A 1 A 2
Between Omsk and Moscow is approximately 2500 km, so the angle A 1 O A 2 is approximately 2500/6400 = 0.4 rad.
Since A 1 A 2 = OA 1 · 0.4, B 1 B 2 = OB 1 · 0.4, and the distance between the upper and lower shelves is about 1 meter, I will drive on A 1 A 2 - B 1 B 2 = ( OA 1 - OB 1 ) · 0.4 = 1 · 0.4 m = 40 centimeters more for the same money.
')
My friend Alexander arranged meetings with three girls, whose names were Nastya, Sveta and Katya, on the same evening. He arranged to meet Nastya next to the house 90 k5 on the Fontanka Embankment, with Sveta at the intersection of Liteiny and Kirochnaya, and with Katya near house 5 on Fontannaya Street (points H , C and K in the figure). Alexander came home from work to meet with Nastya, but halfway through the thought that Sveta was more beautiful, and went to the meeting place with her. Having gone half way to Sveta, he thought that he would have more chances with Katya, and went to where they had agreed to meet. But halfway through, he decided that Nastya was still smarter and went towards Nastya. Where did my friend end up if he continued to walk like that all evening?
Decision
Suppose Sanya is able to walk through walls and other obstacles. Place St. Petersburg on the complex plane and let it work in Z 0 . Then the sequence of points at which he changed the solutions is: Z 1 = ( Z 0 + H ) / 2, Z 2 = ( Z 1 + C ) / 2, Z 3 = ( Z 2 + K ) / 2, Z 4 = ( Z 3 + H ) / 2, ... Divide this sequence into three subsequences: Z 0 , Z 3 , Z 6 , Z 9 , ...; Z 1 , Z 4 , Z 7 , Z 10 , ... and Z 2 , Z 5 , Z 8 , Z 11 , ... Let f ( Z ) = ((( Z + H ) / 2 + C ) / 2 + K ) / 2. Then the first subsequence is written in the form Z 0 , f ( Z 0 ), f ( f ( Z 0 )), f ( f ( f ( Z 0 ))), ... Note that the map f is contractive, therefore, by the Banach fixed point theorem, there is a unique fixed point of this map; we denote it by X. Moreover, this subsequence converges to this point. For the other two subsequences, for similar reasons, there are also fixed points, which we denote by F and T, respectively. These points correspond to the addresses: Nekrasova, 26, Rubinstein, 5 and Belinsky, 11 (see fig.). The first address is the Chronicle bar, the second is Fiddler's Green bar, and the third is Terminal. Thus, Alexander will spend the rest of the day in these bars, moving in turn from one to another.
If you go outside, you can see on the pillars and walls of houses a lot of pieces of paper with pairs, the first element of which is the female name, and the second element is a sequence of numbers. Examples are given below.
Not at all interested in the second elements of these pairs, let's see what the distribution of the first elements is.
For this, I somehow walked six kilometers down the street in August and collected statistics.
It was processed 263 ads, TOP-5 names looks like this: Alice, Lena, Sabina, Dasha, Diana:
My friend Konstantin every evening at 7 o'clock comes home from work and goes to the subway. There he randomly chooses a branch and direction, sits on the train, drives one station, then again randomly selects a branch and direction, again drives one station, and so on, until he reaches home. Knowing that Kostya works for Leo Tolstoy, and lives near art. m. Belyaevo, determine how often he drinks beer, if he buys beer in a store near the house after work.
Decision
The Moscow metro can be associated with a Markov chain with a matrix P (see fig.). And, since Kostya goes to st. m. Belyaevo, then this state B will be absorbing.
(Click on the picture to enlarge.)
Further, suppose that one drive takes about three minutes. Since in Moscow they stop selling beer at 23:00, and (23:00 - 19:00) / 3 min. = 80, then we need to find the probability to be in state B in less than 80 steps. It is easy to understand that in order to find this probability, one needs to raise the matrix P to the power of 79 and take the element at the intersection of the row that corresponds to the Park of Culture and the column that corresponds to Belyaevo. Calculations show (see fig. Below) that the desired probability is approximately equal to 5.9%. Consequently, my friend drinks beer about once every 17 days.
Some time ago a picture walked on the Internet with the fact that Cpshina Pringles is a hyperbolic paraboloid:
When I saw her, I said to myself: “Aha! And since the hyperbolic paraboloid is a ruled surface, the chipsin should crawl through a straight slit. ” Therefore, I bought chips the next day, cut a slot in a carton and did this experiment:
Another example of a ruled surface is a single-cavity hyperboloid. You can check this fact with the help of a bottle of yogurt Miracle. We take a bottle from under the yoghurt, cut out a part of the bottle, which is something like a single-cavity hyperboloid, again cut a slot in the cardboard sheet, insert a piece of the bottle into the slot and twist:
Have a good day!
1. Why do I like to ride the top shelves on the train?
Imagine that the earth is a ball and we are traveling with a friend from Omsk to Moscow along an arc of a circle, then my friend, who drives on the lower shelf, will travel a path equal to the arc length B 1 B 2 , and, I am a path equal to the arc length A 1 A 2
Between Omsk and Moscow is approximately 2500 km, so the angle A 1 O A 2 is approximately 2500/6400 = 0.4 rad.
Since A 1 A 2 = OA 1 · 0.4, B 1 B 2 = OB 1 · 0.4, and the distance between the upper and lower shelves is about 1 meter, I will drive on A 1 A 2 - B 1 B 2 = ( OA 1 - OB 1 ) · 0.4 = 1 · 0.4 m = 40 centimeters more for the same money.
')
2. The task about my friend Sanya
My friend Alexander arranged meetings with three girls, whose names were Nastya, Sveta and Katya, on the same evening. He arranged to meet Nastya next to the house 90 k5 on the Fontanka Embankment, with Sveta at the intersection of Liteiny and Kirochnaya, and with Katya near house 5 on Fontannaya Street (points H , C and K in the figure). Alexander came home from work to meet with Nastya, but halfway through the thought that Sveta was more beautiful, and went to the meeting place with her. Having gone half way to Sveta, he thought that he would have more chances with Katya, and went to where they had agreed to meet. But halfway through, he decided that Nastya was still smarter and went towards Nastya. Where did my friend end up if he continued to walk like that all evening?
Decision
Suppose Sanya is able to walk through walls and other obstacles. Place St. Petersburg on the complex plane and let it work in Z 0 . Then the sequence of points at which he changed the solutions is: Z 1 = ( Z 0 + H ) / 2, Z 2 = ( Z 1 + C ) / 2, Z 3 = ( Z 2 + K ) / 2, Z 4 = ( Z 3 + H ) / 2, ... Divide this sequence into three subsequences: Z 0 , Z 3 , Z 6 , Z 9 , ...; Z 1 , Z 4 , Z 7 , Z 10 , ... and Z 2 , Z 5 , Z 8 , Z 11 , ... Let f ( Z ) = ((( Z + H ) / 2 + C ) / 2 + K ) / 2. Then the first subsequence is written in the form Z 0 , f ( Z 0 ), f ( f ( Z 0 )), f ( f ( f ( Z 0 ))), ... Note that the map f is contractive, therefore, by the Banach fixed point theorem, there is a unique fixed point of this map; we denote it by X. Moreover, this subsequence converges to this point. For the other two subsequences, for similar reasons, there are also fixed points, which we denote by F and T, respectively. These points correspond to the addresses: Nekrasova, 26, Rubinstein, 5 and Belinsky, 11 (see fig.). The first address is the Chronicle bar, the second is Fiddler's Green bar, and the third is Terminal. Thus, Alexander will spend the rest of the day in these bars, moving in turn from one to another.
3. We learn to see data everywhere
If you go outside, you can see on the pillars and walls of houses a lot of pieces of paper with pairs, the first element of which is the female name, and the second element is a sequence of numbers. Examples are given below.
Not at all interested in the second elements of these pairs, let's see what the distribution of the first elements is.
For this, I somehow walked six kilometers down the street in August and collected statistics.
It was processed 263 ads, TOP-5 names looks like this: Alice, Lena, Sabina, Dasha, Diana:
4. The task about my friend Kostya
My friend Konstantin every evening at 7 o'clock comes home from work and goes to the subway. There he randomly chooses a branch and direction, sits on the train, drives one station, then again randomly selects a branch and direction, again drives one station, and so on, until he reaches home. Knowing that Kostya works for Leo Tolstoy, and lives near art. m. Belyaevo, determine how often he drinks beer, if he buys beer in a store near the house after work.
Decision
The Moscow metro can be associated with a Markov chain with a matrix P (see fig.). And, since Kostya goes to st. m. Belyaevo, then this state B will be absorbing.
(Click on the picture to enlarge.)
Further, suppose that one drive takes about three minutes. Since in Moscow they stop selling beer at 23:00, and (23:00 - 19:00) / 3 min. = 80, then we need to find the probability to be in state B in less than 80 steps. It is easy to understand that in order to find this probability, one needs to raise the matrix P to the power of 79 and take the element at the intersection of the row that corresponds to the Park of Culture and the column that corresponds to Belyaevo. Calculations show (see fig. Below) that the desired probability is approximately equal to 5.9%. Consequently, my friend drinks beer about once every 17 days.
5. Chipsoids
Some time ago a picture walked on the Internet with the fact that Cpshina Pringles is a hyperbolic paraboloid:
When I saw her, I said to myself: “Aha! And since the hyperbolic paraboloid is a ruled surface, the chipsin should crawl through a straight slit. ” Therefore, I bought chips the next day, cut a slot in a carton and did this experiment:
Another example of a ruled surface is a single-cavity hyperboloid. You can check this fact with the help of a bottle of yogurt Miracle. We take a bottle from under the yoghurt, cut out a part of the bottle, which is something like a single-cavity hyperboloid, again cut a slot in the cardboard sheet, insert a piece of the bottle into the slot and twist:
Have a good day!
Source: https://habr.com/ru/post/252837/
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