Tasks for movement to prepare for the exam in mathematics (2020). Open event in mathematics How to find the distance knowing the speed of convergence
So let's say our bodies move in the same direction. How many cases do you think there might be for such a condition? That's right, two.
Why is it so? I am sure that after all the examples you will easily figure out how to derive these formulas.
Got it? Well done! It's time to solve the problem.
The fourth task
Kolya goes to work by car at a speed of km/h. Colleague Kolya Vova travels at a speed of km/h. Kolya lives at a distance of km from Vova.
How long will it take Vova to overtake Kolya if they left the house at the same time?
Did you count? Let's compare the answers - it turned out that Vova will catch up with Kolya in hours or minutes.
Let's compare our solutions...
The drawing looks like this:
Similar to yours? Well done!
Since the problem asks how long the guys met and left at the same time, the time they traveled will be the same, as well as the meeting place (in the figure it is indicated by a dot). Making equations, take the time for.
So, Vova made his way to the meeting place. Kolya made his way to the meeting place. This is clear. Now we deal with the axis of movement.
Let's start with the path that Kolya did. Its path () is shown as a segment in the figure. And what does Vova's path () consist of? That's right, from the sum of the segments and, where is the initial distance between the guys, and is equal to the path that Kolya did.
Based on these conclusions, we obtain the equation:
Got it? If not, just read this equation again and look at the points marked on the axis. Drawing helps, doesn't it?
hours or minutes minutes.
I hope that in this example you understand how important the role of well crafted drawing!
And we are smoothly moving on, or rather, we have already moved on to the next step in our algorithm - bringing all quantities to the same dimension.
The rule of three "P" - dimension, reasonableness, calculation.
Dimension.
Not always in tasks the same dimension is given for each participant in the movement (as it was in our easy tasks).
For example, you can meet tasks where it is said that the bodies moved a certain number of minutes, and the speed of their movement is indicated in km / h.
We can't just take and substitute the values in the formula - the answer will be wrong. Even in terms of units of measurement, our answer “will not pass” the test for reasonableness. Compare:
See? With proper multiplication, we also reduce the units of measurement, and, accordingly, we get a reasonable and correct result.
And what happens if we do not translate into one system of measurement? The answer has a strange dimension and % is an incorrect result.
So, just in case, let me remind you the meanings of the basic units of measurement of length and time.
Length units:
centimeter = millimeters
decimeter = centimeters = millimeters
meter = decimeters = centimeters = millimeters
kilometer = meters
Time units:
minute = seconds
hour = minutes = seconds
days = hours = minutes = seconds
Advice: When converting units of measurement related to time (minutes to hours, hours to seconds, etc.), imagine a clock face in your head. It can be seen with the naked eye that minutes is a quarter of the dial, i.e. hours, minutes is a third of the dial, i.e. hours, and a minute is an hour.
And now a very simple task:
Masha rode her bicycle from home to the village at a speed of km/h for minutes. What is the distance between the car house and the village?
Did you count? The correct answer is km.
minutes is an hour, and another minute from an hour (mentally imagined a clock face, and said that minutes is a quarter of an hour), respectively - min \u003d h.
Intelligence.
Do you understand that the speed of a car cannot be km/h, unless, of course, we are talking about a sports car? And even more so, it cannot be negative, right? So, reasonableness, that's about it)
Calculation.
See if your solution "passes" the dimension and reasonableness, and only then check the calculations. It is logical - if there is an inconsistency with dimension and reasonableness, then it is easier to cross out everything and start looking for logical and mathematical errors.
"Love for tables" or "when drawing is not enough"
Far from always, the tasks for movement are as simple as we solved before. Very often, in order to correctly solve a problem, you need to not just draw a competent drawing, but also make a table with all the conditions given to us.
First task
From point to point, the distance between which is km, a cyclist and a motorcyclist left at the same time. It is known that a motorcyclist travels more miles per hour than a cyclist.
Determine the speed of the cyclist if it is known that he arrived at the point a minute later than the motorcyclist.
Here is such a task. Pull yourself together and read it several times. Read? Start drawing - straight line, point, point, two arrows ...
In general, draw, and now let's compare what you got.
Kind of empty, right? We draw a table.
As you remember, all movement tasks consist of components: speed, time and path. It is from these graphs that any table in such problems will consist.
True, we will add one more column - name about whom we write information - a motorcyclist and a cyclist.
Also indicate in the header dimension, in which you will enter the values \u200b\u200bin there. You remember how important this is, right?
Do you have a table like this?
Now let's analyze everything that we have, and in parallel enter the data into a table and into a figure.
The first thing we have is the path that the cyclist and motorcyclist have traveled. It is the same and equal to km. We bring in!
Let us take the speed of the cyclist as, then the speed of the motorcyclist will be ...
If with such variable solution task will not work - it's okay, let's take another one until we reach the victorious one. This happens, the main thing is not to be nervous!
The table has changed. We have left not filled only one column - time. How to find the time when there is a path and speed?
That's right, divide the path by the speed. Enter it in the table.
So our table has been filled, now you can enter data into the figure.
What can we reflect on it?
Well done. The speed of movement of a motorcyclist and a cyclist.
Let's read the problem again, look at the figure and the completed table.
What data is not shown in the table or in the figure?
Right. The time by which the motorcyclist arrived earlier than the cyclist. We know that the time difference is minutes.
What should we do next? That's right, translate the time given to us from minutes to hours, because the speed is given to us in km / h.
The magic of formulas: writing and solving equations - manipulations that lead to the only correct answer.
So, as you already guessed, now we will make up the equation.
Compilation of the equation:
Look at your table, at the last condition that was not included in it, and think about the relationship between what and what can we put into the equation?
Correctly. We can make an equation based on the time difference!
Is it logical? The cyclist rode more, if we subtract the time of the motorcyclist from his time, we will just get the difference given to us.
This equation is rational. If you don't know what it is, read the topic "".
We bring the terms to a common denominator:
Let's open the brackets and give like terms: Phew! Got it? Try your hand at the next task.
Equation solution:
From this equation we get the following:
Let's open the brackets and move everything to left side equations:
Voila! We have a simple quadratic equation. We decide!
We received two responses. Look what we got for? That's right, the speed of the cyclist.
We recall the rule "3P", more specifically "reasonableness". Do you understand what I mean? Exactly! Speed cannot be negative, so our answer is km/h.
Second task
Two cyclists set out on a 1-kilometer run at the same time. The first one was driving at a speed that was 1 km/h faster than the second one, and arrived at the finish line hours earlier than the second one. Find the speed of the cyclist who came to the finish line second. Give your answer in km/h.
I recall the solution algorithm:
- Read the problem a couple of times - learn all the details. Got it?
- Start drawing the drawing - in which direction are they moving? how far did they travel? Did you draw?
- Check if all the quantities you have are of the same dimension and start writing out the condition of the problem briefly, making up a table (do you remember what columns are there?).
- While writing all this, think about what to take for? Chose? Record in the table! Well, now it’s simple: we make an equation and solve it. Yes, and finally - remember the "3P"!
- I've done everything? Well done! It turned out that the speed of the cyclist is km / h.
-"What color is your car?" - "She's beautiful!" Correct answers to the questions
Let's continue our conversation. So what is the speed of the first cyclist? km/h? I really hope you're not nodding in the affirmative right now!
Read the question carefully: "What is the speed of first cyclist?
Got what I mean?
Exactly! Received is not always the answer to the question!
Read the questions carefully - perhaps, after finding it, you will need to perform some more manipulations, for example, add km / h, as in our task.
Another point - often in tasks everything is indicated in hours, and the answer is asked to be expressed in minutes, or all the data is given in km, and the answer is asked to be written in meters.
Look at the dimension not only during the solution itself, but also when writing down the answers.
Tasks for movement in a circle
The bodies in the tasks may not necessarily move in a straight line, but also in a circle, for example, cyclists can ride along a circular track. Let's take a look at this problem.
Task #1
From paragraph circular track the cyclist left. In minutes he had not yet returned to the checkpoint, and a motorcyclist followed him from the checkpoint. Minutes after departure, he caught up with the cyclist for the first time, and minutes after that he caught up with him for the second time.
Find the speed of the cyclist if the length of the track is km. Give your answer in km/h.
Solution of problem No. 1
Try to draw a picture for this problem and fill in the table for it. Here's what happened to me:
Between meetings, the cyclist traveled the distance, and the motorcyclist -.
But at the same time, the motorcyclist drove exactly one lap more, this can be seen from the figure:
I hope you understand that they didn't actually go in a spiral - the spiral just schematically shows that they go in a circle, passing the same points of the track several times.
Got it? Try to solve the following problems yourself:
Tasks for independent work:
- Two mo-to-tsik-li-hundreds start-to-tu-yut one-but-time-men-but in one-right-le-ni from two dia-met-ral-but pro-ty-in-po- false points of a circular route, the length of a swarm is equal to km. After how many minutes, mo-the-cycle-lists are equal for the first time, if the speed of one of them is by km / h more than the speed of the other th?
- From one point of the circle-howl of the highway, the length of some swarm is equal to km, at the same time, in one right-le-ni, there are two motorcyclists. The speed of the first motorcycle is km / h, and minutes after the start, he was ahead of the second motorcycle by one lap. Find the speed of the second motorcycle. Give your answer in km/h.
Solving problems for independent work:
- Let km / h be the speed of the first mo-to-cycle-li-hundred, then the speed of the second mo-to-cycle-li-hundred is km / h. Let the first time mo-the-cycle-lists be equal in hours. In order for mo-the-cycle-li-stas to be equal, the faster one must overcome them from the beginning distance, equal in lo-vi-not to the length of the route.
We get that the time is equal to hours = minutes.
- Let the speed of the second motorcycle be km/h. In an hour, the first motorcycle traveled a kilometer more than the second swarm, respectively, we get the equation:
The speed of the second motorcyclist is km/h.
Tasks for the course
Now that you're good at solving problems "on land", let's move on to the water and look at the scary problems associated with the current.
Imagine that you have a raft and you lower it into a lake. What is happening to him? Correctly. It stands because a lake, a pond, a puddle, after all, is stagnant water.
The current velocity in the lake is .
The raft will only move if you start rowing yourself. The speed he gains will be own speed of the raft. No matter where you swim - left, right, the raft will move at the same speed with which you row. This is clear? It's logical.
Now imagine that you are lowering the raft onto the river, turn away to take the rope ..., turn around, and he ... floated away ...
This happens because the river has a flow rate, which carries your raft in the direction of the current.
At the same time, its speed is equal to zero (you are standing in shock on the shore and not rowing) - it moves with the speed of the current.
Got it?
Then answer this question - "How fast will the raft float on the river if you sit and row?" Thinking?
Two options are possible here.
Option 1 - you go with the flow.
And then you swim at your own speed + the speed of the current. The current seems to help you move forward.
2nd option - t You are swimming against the current.
Hard? That's right, because the current is trying to "throw" you back. You make more and more efforts to swim at least meters, respectively, the speed with which you move is equal to your own speed - the speed of the current.
Let's say you need to swim a mile. When will you cover this distance faster? When will you move with the flow or against?
Let's solve the problem and check.
Let's add to our path data on the speed of the current - km/h and on the own speed of the raft - km/h. How much time will you spend moving with and against the current?
Of course, you easily coped with this task! Downstream - an hour, and against the current as much as an hour!
This is the whole essence of the tasks on flow with the flow.
Let's complicate the task a little.
Task #1
A boat with a motor sailed from point to point in an hour, and back in an hour.
Find the speed of the current if the speed of the boat is standing water- km/h
Solution of problem No. 1
Let's denote the distance between the points as, and the speed of the current as.
| Path S | speed v, km/h |
time t, hours |
|
| A -> B (upstream) | 3 | ||
| B -> A (downstream) | 2 |
We see that the boat makes the same path, respectively:
What did we charge for?
Flow speed. Then this will be the answer :)
The speed of the current is km/h.
Task #2
The kayak went from point to point, located km away. After staying at point for an hour, the kayak set off and returned to point c.
Determine (in km/h) the own speed of the kayak if it is known that the speed of the river is km/h.
Solution of problem No. 2
So let's get started. Read the problem several times and draw a picture. I think you can easily solve this on your own.
Are all quantities expressed in the same form? No. The rest time is indicated both in hours and in minutes.
Converting this to hours:
hour minutes = h.
Now all quantities are expressed in one form. Let's start filling out the table and looking for what we'll take for.
Let be the own speed of the kayak. Then, the speed of the kayak downstream is equal, and against the current is equal.
Let's write this data, as well as the path (as you understand, it is the same) and the time expressed in terms of path and speed, in a table:
| Path S | speed v, km/h |
time t, hours |
|
| Against the stream | 26 | ||
| With the flow | 26 |
Let's calculate how much time the kayak spent on its trip:
Did she swim all hours? Rereading the task.
No, not all. She had a rest of an hour of minutes, respectively, from the hours we subtract the rest time, which we have already translated into hours:
h kayak really floated.
Let's bring all the terms to a common denominator:
We open the brackets and give like terms. Next, we solve the resulting quadratic equation.
With this, I think you can also handle it on your own. What answer did you get? I have km/h.
Summing up
ADVANCED LEVEL
Movement tasks. Examples
Consider examples with solutionsfor each type of task.
moving with the flow
One of the simplest tasks tasks for the movement on the river. Their whole essence is as follows:
- if we move with the flow, the speed of the current is added to our speed;
- if we move against the current, the speed of the current is subtracted from our speed.
Example #1:
The boat sailed from point A to point B in hours and back in hours. Find the speed of the current if the speed of the boat in still water is km/h.
Solution #1:
Let's denote the distance between the points as AB, and the speed of the current as.
We will enter all the data from the condition in the table:
| Path S | speed v, km/h |
Time t, hours | |
| A -> B (upstream) | AB | 50s | 5 |
| B -> A (downstream) | AB | 50+x | 3 |
For each row of this table, you need to write the formula:
In fact, you don't have to write equations for each of the rows in the table. We see that the distance traveled by the boat back and forth is the same.
So we can equate the distance. To do this, we immediately use distance formula:
Often it is necessary to use formula for time:
Example #2:
A boat travels a distance in km against the current for an hour longer than with the current. Find the speed of the boat in still water if the speed of the current is km/h.
Solution #2:
Let's try to write an equation. The time upstream is one hour longer than the time downstream.
It is written like this:
Now, instead of each time, we substitute the formula:
Got the usual rational equation, let's solve it:
Obviously, speed cannot be a negative number, so the answer is km/h.
Relative motion
If some bodies are moving relative to each other, it is often useful to calculate their relative speed. It is equal to:
- the sum of velocities if the bodies move towards each other;
- speed difference if the bodies are moving in the same direction.
Example #1
From points A and B, two cars left simultaneously towards each other with speeds of km/h and km/h. In how many minutes will they meet? If the distance between points is km?
I solution way:
Relative speed of cars km/h. This means that if we are sitting in the first car, it seems to be stationary, but the second car is approaching us at a speed of km/h. Since the distance between cars is initially km, the time after which the second car will pass the first:
Solution 2:
The time from the start of the movement to the meeting at the cars is obviously the same. Let's designate it. Then the first car drove the way, and the second -.
In total, they traveled all km. Means,
Other motion tasks
Example #1:
A car left point A for point B. Simultaneously with it, another car left, which traveled exactly half the way at a speed of km/h less than the first, and the second half of the way it drove at a speed of km/h.
As a result, the cars arrived at point B at the same time.
Find the speed of the first car if it is known to be greater than km/h.
Solution #1:
To the left of the equal sign, we write the time of the first car, and to the right - the second:
Simplify the expression on the right side:
We divide each term by AB:
It turned out the usual rational equation. Solving it, we get two roots:
Of these, only one is larger.
Answer: km/h.
Example #2
A cyclist left point A of the circular track. After a few minutes, he had not yet returned to point A, and a motorcyclist followed him from point A. Minutes after departure, he caught up with the cyclist for the first time, and minutes after that he caught up with him for the second time. Find the speed of the cyclist if the length of the track is km. Give your answer in km/h.
Solution:
Here we will equate the distance.
Let the speed of the cyclist be, and the speed of the motorcyclist -. Until the moment of the first meeting, the cyclist was on the road for minutes, and the motorcyclist -.
In doing so, they traveled equal distances:
Between meetings, the cyclist traveled the distance, and the motorcyclist -. But at the same time, the motorcyclist drove exactly one lap more, this can be seen from the figure:
I hope you understand that they didn't actually go in a spiral - the spiral just schematically shows that they go in a circle, passing the same points of the track several times.
We solve the resulting equations in the system:
SUMMARY AND BASIC FORMULA
1. Basic formula
2. Relative motion
- This is the sum of the speeds if the bodies are moving towards each other;
- speed difference if the bodies are moving in the same direction.
3. Move with the flow:
- If we move with the current, the speed of the current is added to our speed;
- if we move against the current, the speed of the current is subtracted from the speed.
We have helped you deal with the tasks of movement...
Now it's your turn...
If you carefully read the text and solved all the examples yourself, we are ready to argue that you understood everything.
And this is already half way.
Write below in the comments if you figured out the tasks for movement?
Which cause the greatest difficulty?
Do you understand that tasks for "work" are almost the same thing?
Write to us and good luck on your exams!
Let the motion of the first body be characterized by the values s 1 , v 1 , t 1 , and the motion of the second body - s 2 , v 2 , t 2 . Such a movement can be represented in a schematic drawing: v 1, t 1 t built-in. v2, t2
If two objects start moving towards each other at the same time, then each of them spends the same time from the moment of movement until the meeting - meeting time, i.e. t 1= t 2= t int.
The distance that moving objects approach each other per unit of time is called approach speed, those. v bl. \u003d v 1 + v 2.
The distance between the bodies can be expressed as follows: s=s 1 +s 2.
The entire distance traveled by moving bodies in oncoming traffic can be calculated by the formula: s=v sbl. t int. .
Example. Let's solve the problem: "Two pedestrians simultaneously came out towards each other from two points, the distance between which is 18 km. The speed of one of them is 5 km / h, the other - 4 km / h. In how many hours will they meet?
Solution: The problem considers the movement towards the meeting of two pedestrians. One travels at 5 km/h, the other at 4 km/h. The path they have to go is 18 km. It is required to find the time after which they will meet, starting to move at the same time.
| Members of the movement | Speed | Time | Distance |
| First pedestrian | 5km/h | ?h - the same | 18 km |
| Second pedestrian | 4km/h |
Since the speeds of pedestrians are known, we can find their approach speed: 5+4=9(km/h). Then, knowing the speed of approach and the distance they need to go, you can find the time after which the pedestrians will meet: 189=2(h).
Problems for the movement of two bodies in the same direction.
Two types of such tasks are distinguished: 1) the movement starts simultaneously from different points; 2) the movement starts at the time from one point.
Let the motion of the first body be characterized by the values s 1 , v 1 , t 1 , and the motion of the second body - s 2 , v 2 , t 2 . Such a movement can be represented in a schematic drawing:
v 1 , t 1 v 2 , t 2 t
If, when moving in one direction, the first body catches up with the second, then v 1 v 2, in addition, per unit time, the first object approaches the other at a distance v 1 -v 2 . This distance is called approach speed: v sbl. =v 1 -v 2 .
The distance between the bodies can be expressed by the formulas: s= s 1 - s 2 and s= v sbl. t int.
Example. Let's solve the problem: "From two points, remote from each other at a distance of 30 km. The speed of one is 40 km/h, the other 50 km/h. In how many hours will the second rider overtake the first?
Solution: The problem considers the movement of two motorcyclists. They left at the same time from different points located at a distance of 30 km. The speed of one is 40 km / h, the other is 50 km / h. It is required to find out in how many hours the second motorcyclist will catch up with the first.
Auxiliary models can be different - a schematic drawing (see above) and a table:
Knowing the speed of both motorcyclists, you can find out their approach speed: 50-40=10(km/h). Then, knowing the speed of approach and the distance between motorcyclists, we will find the time during which the second motorcyclist will overtake the first one: 3010=3(h).
Let us give an example of a problem that describes the second situation of motion of two bodies in the same direction.
Example. Let's solve the problem: “A train left Moscow at 7 o'clock at a speed of 60 km/h. At 13:00 the next day, a plane took off in the same direction at a speed of 780 km/h. How long will it take for the plane to overtake the train?
Solution: The problem considers the movement of a train and an airplane in the same direction from the same point, but at different times. It is known that the speed of the train is 60 km/h, the speed of the aircraft is 780 km/h; the train starts at 7:00 and the plane starts at 13:00 the next day. It is required to find out how long it will take for the plane to overtake the train.
It follows from the conditions of the problem that by the time the plane takes off, the train has traveled a certain distance. If you find it, then given task becomes similar to the previous problem.
To find this distance, you need to calculate how long the train was on the way: 24-7 + 13 = 30 (h). Knowing the speed of the train and the time it was on the way before the departure of the plane, you can find the distance between the train and the plane: 6030=1800(km). Then we find the speed of approach of the train and the plane: 780-60=720(km/h). And further, the time after which the plane will catch up with the train: 1800720=2.5(h).
math journey
Here are the ideas and tasks,
Games, jokes, everything for you!
We wish you good luck
To work, have a good time!
To the gray heron for a lesson Arrived 7 forty, And they have only 3 magpies prepared lessons. How many loafers-forty Arrived at the lesson?
They gave the children a lesson at school: Jumping in the field 40 forty, Ten took off Sat on the fir. How many left in the field of forty?
We are a big family
Most junior is me.
Do not immediately count us:
Manya is and Vanya is,
Yura, Shura, Klasha, Sasha
And Natasha is also ours.
We're walking down the street
They say it's an orphanage.
Count quickly
How many of us are children in the family.
Mom will allow today
After school I go for a walk.
I am not too much and not too little
Got marked...
There is a long segment, there is a shorter one,
By the ruler we draw it, by the way.
Centimeters five - size,
It's called...
It consists of a point and a line.
Well, guess who he is?
It happens that in the rain it breaks through the clouds.
Now guess what? It...
If two objects are far apart,
We can easily calculate the kilometers between them.
Speed, time - we know the values,
We now multiply their values.
The result of all our knowledge -
Counted...
He is bipedal but lame
Draws with only one foot.
Stand in the center with the second leg,
So that the circle of the curve does not come out.
Metagrams
A certain word is encrypted in a metagram. It needs to be guessed. Then, in the deciphered word, one of the indicated letters should be replaced by another letter, and the meaning of the word will change.
He is not a very small rodent,
For a little more squirrels.
And replace "U" with "O" -
It will be a round number.
Answer: With at rock - with about rock.
With "Sh" - I need to count,
With "M" - offenders are terrible!
Answer: w there is - m there is
info-know-it-all
Now let everyone know Who is the smartest? Who is more well-read, wiser - Win this contest!
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AWARDING
THANK YOU ALL! YOU ARE GREAT!
The most difficult and least formalized in the task of automatic classification is the moment associated with the definition of the concept of homogeneity of objects.
In the general case, the concept of homogeneity of objects is determined by setting the rule for calculating the value characterizing either the distance between objects from the studied population or the degree of proximity (similarity) of the same objects. If the function is given, then objects that are close in the sense of this metric are considered to be homogeneous, belonging to the same class. Naturally, this requires a comparison with a certain threshold value determined in each specific case in my own way.
Similarly, the above-mentioned proximity measure is used to form homogeneous classes, when specifying which one must remember the need to comply with the following natural requirements: the symmetry requirements of the requirement for the maximum similarity of the object with itself and the requirements for a given metric of monotonous decrease in , i.e. from must necessarily follow fulfillment of the inequality
Of course, the choice of a metric (or proximity measure) is the key moment of the study, on which the final version of the division of objects into classes depends decisively for a given partitioning algorithm. In each specific task, this choice should be made in its own way. At the same time, the solution of this issue depends mainly on the main objectives of the study, the physical and statistical nature of the observation vector X, the completeness of a priori information about the nature of the probability distribution of X. For example, if it follows from the final objectives of the study and from the nature of the vector X that the concept of a homogeneous group it is natural to interpret as a general population with a single-vertex density (frequency polygon) distribution, and if, in addition, the general form of this density is known, then the general approach described in Chap. 6. If, in addition, it is known that observations are drawn from normal populations with the same covariance matrix, then a natural measure of the distance between two objects from each other is the distance of the Mahalanobis type (see below).
As examples of distances and proximity measures that are relatively widely used in cluster analysis problems, we present the following.
General view of the Mahalanobis type metric. In the general case of the dependent components of the observation vector X and their different significance, in deciding whether an object (observation) belongs to a particular class, they usually use the generalized ("weighted") distance of the Mahalanobis type, given by the formula
Here is the covariance matrix of the general population from which observations are extracted, and A is some symmetric non-negative definite matrix of "weight" coefficients, which is most often chosen to be diagonal.
The next three types of distances, although they are special cases of the metric, still deserve a special description.
Common Euclidean distance
Situations in which the use of this distance can be considered justified primarily include the following:
observations X are extracted from the populations described by the multivariate normal law With covariance matrix of the form, i.e., the components of X are mutually independent and have the same variance;
the components of the observation vector X are homogeneous in their physical meaning, and it was established, for example, with the help of a survey of experts, that all of them are equally important from the point of view of resolving the issue of attributing an object to a particular class;
the attribute space coincides with the geometric space of our being, which can only be in cases , and the concept of the proximity of objects, respectively, coincides with the concept of geometric proximity in this space, for example, the classification of hits when shooting at a target.
"Weighted" Euclidean distance
It is usually used in situations in which one way or another it is possible to assign some non-negative "weight" to each of the components of the observation vector X.
The definition of weights is usually associated with additional research, for example, obtaining and using training samples, organizing a survey of experts and processing their opinions, using some special models. Attempts to determine the weights only from the information contained in the initial data, as a rule, do not give the desired effect, and sometimes can only move away from the true solution. It suffices to note that, depending on the very subtle and insignificant variations in the physical and statistical nature of the initial data, equally convincing arguments can be made in favor of two diametrically opposite solutions to this issue - to choose in proportion to the value of the mean square error of the feature or in proportion to the reciprocal of the mean square error of the same feature.
Hamming distance. It is used as a measure of the difference between objects defined by dichotomous features. It is given using the formula
and, therefore, is equal to the number of mismatches in the values of the corresponding features in the objects under consideration.
Other proximity measures for dichotomous features.
Measures of proximity of objects described by a set of dichotomous features are usually based on the characteristics can be considered equal, and the effect of the coincidence or mismatch of zeros is the same as that of the coincidence or mismatch of ones, then d as a measure of the proximity of objects use the value
A very complete overview of the various measures of proximity of objects described by dichotomous features, the reader will find in.
Proximity and distance measures specified using a potential function. In many problems of mathematical statistics, probability theory, physical theory potential and the theory of pattern recognition, or the classification of multivariate observations, some specially arranged functions of two vector variables X and Y, and most often simply of the distance between these variables, which we will call potential, turn out to be useful.
So, for example, if the space of all conceivable values of the investigated vector X is divided into a complete system of disjoint simply connected compact sets or homogeneous classes and the potential function is defined for as follows:
Otherwise, using this function it is convenient to construct ordinary empirical histograms (estimates of the distribution density from the available observations. Indeed, it is easy to see that
where - the number of observations that fall into the class containing the point - the volume of the region (the geometric interpretation for the one-dimensional case is shown in Fig. 5.1).
If the metric is given in the factor space under study, then we can not bind ourselves by a pre-fixed division into classes, but can be set as a monotonically decreasing function of the distance .
For example,
Here we give just one more general form connection between , in which the distance acts as a function of some values of the potential function K:
Rice. 5.1, Histogram constructed using grouping of a one-dimensional sample population
In particular, by choosing scalar product vectors U and V, i.e., setting
we obtain by formula (5.3) the usual Euclidean distance .
It is easy to understand that even if the potential function is given in the form of relations (5.2), formulas (5.1) make it possible to construct statistical estimates for the distribution density (5.1), although the graph of the function will no longer be stepwise, but smoothed. In the absence of a metric in space, functions can be used as a measure of the proximity of objects u and V, as well as objects and entire classes and classes to each other.
In the first case, this measure made it possible to obtain only a qualitative answer: the objects are close if U and V belong to the same class, and the objects are far away otherwise; in the other two cases, the proximity measure is a quantitative characteristic.
On physically meaningful measures of the proximity of objects. In some problems of classification of objects, which are not necessarily described quantitatively, it is more natural to use as a measure of the proximity of objects (or the distance between them) some physically meaningful numerical parameters, one way or another characterizing the relationship between objects. An example is the classification problem for the purpose of aggregating sectors of the national economy, which is solved on the basis of the input-output matrix. Thus, the classified object in this example is a branch of the national economy, and the input-output matrix is represented by elements where by means the amount of annual supplies in monetary terms of the industry in . As a proximity matrix in this case, it is natural to take, for example, a symmetrized normalized input-output matrix. At the same time, normalization is understood as a transformation in which the monetary value of supplies from an industry to is replaced by the share of these supplies in relation to all supplies of the industry. Symmetrization of the normalized input-output matrix can be carried out in various ways. So, for example, the proximity between industries is expressed either through the average value of their mutual rationed deliveries, or through a combination of their mutual rationed deliveries.
On measures of proximity of numerical features (individual factors). The solution of problems of classification of multidimensional data, as a rule, provides, as a preliminary stage of the study, the implementation of methods that allow to significantly reduce the dimension of the original factor space, to select from the components of the observed vectors X a relatively small number of the most significant, most informative. For these purposes, it is useful to consider each of the components as an object to be classified. The fact is that the division of features into a small number of groups that are homogeneous in a certain sense will allow the researcher to conclude that the components included in one group are, in a certain sense, strongly related to each other and carry information about some one property of the object under study.
Therefore, it can be hoped that there will be no great loss in information if we leave only one representative from each such group for further research.
Most often, in such situations, various characteristics of the degree of their correlation and, first of all, correlation coefficients are used as measures of proximity between individual features, as well as between sets of such features. The problem of reducing the dimension of the analyzed feature space is specially devoted to section III books. In more detail, the issues of constructing and using distances and proximity measures between individual objects are considered in.
Sections, straight
To hell with her quickly!
Fields without difficulty
Will guide you ... (ruler)
Three sides and three corners.
And every student knows:
The figure is called
Of course... (triangle)
To get the amount
You need two numbers... (add)
If we take something
Numbers, children, ... (subtract)
If more than five times,
We will number ... (multiply)
If less, then
We will be the numbers ... (divide)
If it gets into the diary -
The student is at fault:
Long nose, one leg
Like Grandma Yaga.
Spoils a page in the diary
All mark ... ("unit")
Long nose like a bird's beak
This is a number ... ("one")
Kolami, which is in my notebook,
I'll build a fence in the garden.
I'm getting them craftswoman,
My mark... ("one")
For this mark will be
At home, I'm a headache.
I'll tell you a secret:
The number with the letter "3" is similar,
Like twins, look.
You can even confuse
Letter "3" and the number ... ("three")
So many legs on the table
And the corners in the apartment
Guess what, kids?
They are always ... (four)
Marks are better not to find!
"Excellent" means... ("five")
Mom will allow today
After school I go for a walk.
I am not too much and not too little
Received a mark ... ("five")
The figure has a hooked head,
And even the belly is.
The hook looks like a cap
Crossbar along the body
Put on the number.
The kerchief flutters in the wind.
So similar to matryoshka -
Trunk with head.
- What is the number? Let's ask right away.
- Well, of course, the number ... ("eight")
Appeared suddenly in a notebook
"Six" on the head - ... (nine)
He thinks he's a king
But in fact - ... (zero)
She has nothing:
No eyes, no hands, no nose,
It consists of everything
The whole world knows this:
Angle measures ... (protractor)
A task where you need to think.
I'm a student anywhere
I never spoil
Even though I'm not a pioneer
But to all the guys ... (example)
I did it in my notebook
As clear as rhythm
Action after another.
This is... (algorithm)
I am very diligent
Completed ... (task)
These signs are only in pairs,
Round, square.
We see them all the time
We write many times.
We conclude, as in boxes,
Numbers in... (brackets)
This is the magnitude.
And she's the only one
The size of the surfaces measures,
In grams, kilograms too
We can measure it. (Weight)
Five centimeters - the value,
It's called... (length)
Mathematics lesson.
The bell just rang
We are at the desks, and here
Let's start oral ... (account)
Someone needs to explain
What is an hour? Minute?
Since ancient times, any tribe
Knows what is ... (time)
He connects the point of the circle
With its center - everyone knows that.
It is denoted by the letter "g".
Unknown X, unknown Y,
Maybe "minus" - all the same.
Adding, subtracting,
So... we decide. (examples)
You need to know these signs.
Ten of them, but these signs
arithmetic operation,
The reverse of addition,
I will tell you without a doubt.
As a result, the difference
No wonder my efforts!
I solved the example correctly
And this... (subtraction)
Add plus numbers
And then we count the answer.
This action is... (addition)
Movement speed
Consonant with the word "acceleration".
Answer, children, to me now,
Speed, time - we know the quantities,
The result of all our knowledge -
Counted ... (distance)
I go and repeat
And again I remember:
Two by two is four,
Five three is fifteen.
To remember everything
You have to try.
This achievement is... (multiplication table)
He is bipedal but lame
Draws with only one foot.
Stand in the center with the second leg,
It has four sides
All are equal among themselves.
With a rectangle, he is a brother,
It's called... (square)
Compass, our reliable friend,
If fingers are missing
My friends will count me.
I will lay them out on the desk,
Wherever you take her
This line is like this
Without end and without beginning
It's called... (direct)
It is limited on both sides.
And carried out along the line.
You can measure its length
Every kid knows:
The addition sign is ... ("plus")
It consists of a point and a line.
And we can tell you now
What 60 minutes is... (an hour)
The triangle has three
But there are four of them in a square.
He is deployed
Sharp, maybe blunt.
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"Mathematical riddles."
Riddles about mathematical accessories, about signs mathematical operations, riddles about geometric shapes, riddles for children from 9 to 12 years old. Riddles for schoolchildren.
Sections, straight
To hell with her quickly!
Fields without difficulty
Will guide you ... (ruler)
Three sides and three corners.
And every student knows:
The figure is called
Of course... (triangle)
To get the amount
You need two numbers... (add)
If we take something
Numbers, children, ... (subtract)
If more than five times,
We will number ... (multiply)
If less, then
We will be the numbers ... (divide)
If it gets into the diary -
The student is at fault:
Long nose, one leg
Like Grandma Yaga.
Spoils a page in the diary
All mark ... ("unit")
Long nose like a bird's beak
This is a number ... ("one")
Kolami, which is in my notebook,
I'll build a fence in the garden.
I'm getting them craftswoman,
My mark... ("one")
For this mark will be
At home, I'm a headache.
I'll tell you a secret:
Received in a notebook ... ("deuce")
The number with the letter "3" is similar,
Like twins, look.
You can even confuse
Letter "3" and the number ... ("three")
So many legs on the table
And the corners in the apartment
Guess what, kids?
They are always ... (four)
Marks are better not to find!
"Excellent" means... ("five")
Mom will allow today
After school I go for a walk.
I am not too much and not too little
Received a mark ... ("five")
The figure has a head - a hook,
And even the belly is.
The hook looks like a cap
And this figure ... ("six")
Yandex.Direct
Crossbar along the body
Put on the number.
The kerchief flutters in the wind.
What do you say the number is called? ("Seven")
So similar to matryoshka -
Trunk with head.
What's the number? We'll ask right away.
Well, of course, the number ... ("eight")
Appeared suddenly in a notebook
"Six" on the head - ... (nine)
He thinks he's a king
But in fact - ... (zero)
She has nothing:
No eyes, no hands, no nose,
It consists of everything
From the condition with the question. (A task)
The whole world knows this:
Angle measures ... (protractor)
A task where you need to think.
It may not need to be addressed.
What is needed here is not knowledge, but ingenuity,
And the cheat sheet will not help in solving.
If suddenly a breakdown happens in the mind,
Unresolved will remain ... (puzzle)
I'm a student anywhere
I never spoil
Even though I'm not a pioneer
But to all the guys ... (example)
I did it in my notebook
As clear as rhythm
Action after another.
This is... (algorithm)
I am very diligent
Completed ... (task)
These signs are only in pairs,
Round, square.
We see them all the time
We write many times.
We conclude, as in boxes,
Numbers in... (brackets)
This is the magnitude.
And she's the only one
The size of the surfaces measures,
In the square defines. (Square)
In grams, kilograms too
We can measure it. (Weight)
There is a long segment, there is a shorter one,
By the ruler we draw it, by the way.
Centimeters five - size,
It's called... (length)
Mathematics lesson.
The bell just rang
We are at the desks, and here
Let's start oral ... (account)
Someone needs to explain
What is an hour? Minute?
Since ancient times, any tribe
Knows what is ... (time)
He connects the point of the circle
With its center - everyone knows that.
It is denoted by the letter "g".
Can you tell me what it's called? (Circle radius)
Unknown X, unknown Y,
They can be met in equalities.
And this, guys, I'll tell you, it's not a game,
This is where a serious solution needs to be found.
With unknown equals without a doubt
We call guys, how are we? (Equations)
Three plus three and five plus five
There is a plus sign and an equals sign
Maybe "minus" - all the same.
Adding, subtracting,
So... we decide. (examples)
You need to know these signs.
Ten of them, but these signs
They count everything in the world. (numbers)
arithmetic operation,
The reverse of addition,
The minus sign is involved in it,
I will tell you without a doubt.
As a result, the difference
No wonder my efforts!
I solved the example correctly
And this... (subtraction)
In Latin, this word "less" means,
And with us, this sign of the number subtracts. (Minus)
Add plus numbers
And then we count the answer.
If "plus", then, without a doubt,
This action is... (addition)
Movement speed
Consonant with the word "acceleration".
Answer, children, to me now,
What does 8 meters per hour mean? (Speed)
If two objects are far apart,
We can easily calculate the kilometers between them.
Speed, time - we know the values,
We now multiply their values.
The result of all our knowledge -
Counted ... (distance)
I go and repeat
And again I remember:
Two by two is four,
Five three - fifteen.
To remember everything
You have to try.
This achievement is... (multiplication table)
He is bipedal but lame
Draws with only one foot.
Stand in the center with the second leg,
So that the circle of the curve does not come out. (Compass)
Body capacity, part of space
What do we call? I see then... (volume)
It has four sides
All are equal among themselves.
With a rectangle, he is a brother,
It's called... (square)
Compass, our reliable friend,
Draws in a notebook again ... (circle)
One, two, three, four, five...
If fingers are missing
My friends will count me.
I will lay them out on the desk,
And I will solve any example. (Counting sticks)
Wherever you take her
This line is like this
Without end and without beginning
It's called... (direct)
It is limited on both sides.
And carried out along the line.
You can measure its length
And it's so easy to do it! (Line segment)
Every kid knows:
The addition sign is ... ("plus")
It consists of a point and a line.
Well, guess who he is?
It happens that in the rain it breaks through the clouds.
Now guess what? This is... (beam)
We studied time in mathematics,
Everyone-everything-everyone learned about minutes and seconds.
And we can tell you now
What 60 minutes is... (an hour)
The triangle has three
But there are four of them in a square.
All squares are equal to each other.
What am I talking about guys? (Parties)
He is deployed
Sharp, maybe blunt.
What do guys call two beams,
Coming from a point from one? (Corner)
