One of the most famous non-elementary functions, which is used in mathematics, in the theory of differential equations, in statistics and in probability theory, is the Laplace function. Solving problems with it requires significant preparation. Let's find out how you can use Excel tools to calculate this indicator.

The Laplace function has wide applied and theoretical applications. For example, it is quite often used to solve differential equations. This term has another equivalent name - the probability integral. In some cases, the basis for the solution is the construction of a table of values.

Operator NORM.ST.DIST

In Excel, the specified task is solved using the operator STANDARD DIST... Its name is short for Normal Standard Distribution. Since its main task is to return the standard normal cumulative distribution to the selected cell. This operator belongs to the statistical category of standard Excel functions.

In Excel 2007 and in earlier versions of the program, this operator was called NORMSTRASP... For compatibility reasons, it has been retained in modern versions of applications. But still, it is recommended to use a more advanced analogue in them - STANDARD DIST.

Operator syntax STANDARD DIST as follows:

STANDARD ST.DIST (z; cumulative)

Deprecated operator NORMSTRASP written like this:

NORMSTRASP (z)

As you can see, in the new version to the existing argument "Z" added argument "Integral"... It should be noted that each argument is required.

Argument "Z" indicates the numeric value for which the distribution is plotted.

Argument "Integral" represents a boolean value that can be represented "TRUE" ("1") or "LYING" («0») ... In the first case, the cumulative distribution function is returned to the specified cell, and in the second, the weight distribution function.

The solution of the problem

In order to perform the required calculation for a variable, the following formula is applied:

STANDARD DIST (z; cumulative (1)) - 0.5

Now let's look at the use of the operator STANDARD DIST to solve a specific problem.

The Laplace function is a non-elementary function and is often used both in the theory of differential equations and probability theory, and in statistics. The Laplace function requires a certain set of knowledge and training, because it allows you to solve various problems in the field of applied and theoretical application.

The Laplace function is often used to solve differential equations and is often referred to as the integral of probability. Let's see how this function can be used in Excel and how it works.

The operator "NORMSTRASP" corresponds to the integral of probability or the Laplace function in Excel, which has the syntax: "= NORMSTRASP (z). In newer versions of the program, the operator also has the name "NORM.ST.DIST." and a slightly modified syntax “= NORM.ST.DIST (z; cumulative).

The "Z" argument is responsible for the numerical value of the distribution. Argument "Cumulative" - ​​returns two values ​​- "1" - cumulative distribution function, "0" - weight distribution function.

With the theory sorted out. Let's move on to practice. Consider using the Laplace function in Excel.

1. Let's write the value into the cell, insert the function into the next one.

2. Let's write down the function manually “= NORM.ST.DIST (B4; 1).

3. Or we will use the function insertion wizard - go to the “Static” category and indicate the “Complete alphabetical list.

4. In the appeared window of function arguments, point to the initial values. For the variable "Z" our original cell will be responsible, and in the "Integral" we will insert "1". Our function will return the cumulative distribution function.

5. We get a ready-made solution of the standard normal cumulative distribution for the given function "NORM.ST.DIST". But that's not all, our goal was to find the Laplace function or the integral of the probability, so let's take a few more steps.

6. The Laplace function implies that "0.5" must be subtracted from the value of the obtained function. We add the necessary operation to the function. Press "Enter" and get the final solution. The desired value is correct and quickly found.

Excel can easily calculate this function for any cell value, range of cells, or cell references. The "NORM.ST.DIST" function is a standard operator for finding the probability integral or, as it is also called, the Laplace function.

2.1. Laplace function (probability integral) looks like:

The graph of the Laplace function is shown in Fig. 5.

Function F(NS) is tabulated (see Table 1 of the Appendix). To apply this table, you need to know properties of the Laplace function:

1) Function Ф ( NS) odd: F(-NS)= -F(NS).

2) Function F(NS) monotonically increasing.

3) F(0)=0.

4) F(+¥ )=0,5; F(-¥ ) = - 0.5. In practice, we can assume that for x³5 the function F(NS) = 0.5; for x £ -5 the function F(NS)=-0,5.

2.2. There are other forms of the Laplace function:

and

Unlike these forms, the function F(NS) is called the standard or normalized Laplace function. It is associated with other forms of relationships:

EXAMPLE 2. Continuous random variable NS has a normal distribution with parameters: m=3, s= 4. Find the probability that, as a result of the test, a random variable NS: a) will take the value enclosed in the interval (2; 6); b) will take a value less than 2; c) will take a value greater than 10; d) deviates from the mathematical expectation by an amount not exceeding 2. Illustrate the solution of the problem graphically.

Solution. a) The probability that a normal random variable NS falls into the specified interval ( a, b), where a= 2 and b= 6 is equal to:

Laplace function values F (x) determined according to the table given in the appendix, taking into account that F(–NS)= –F(NS).

b) The probability that a normal random variable NS takes a value less than 2, is equal to:

c) The probability that a normal random variable NS takes a value greater than 10, is equal to:

d) The probability that a normal random variable NS d= 2, is equal to:

Geometrically, the calculated probabilities are numerically equal to the shaded areas under the normal curve (see Fig. 6).

Rice. 6. Normal curve for a random variable NS~N(3;4)
EXAMPLE 3. The shaft diameter is measured without systematic (one sign) errors. Random measurement errors are subject to the normal distribution law with a standard deviation of 10 mm. Find the probability that the measurement will be made with an error not exceeding 15 mm in absolute value.

Solution. The mathematical expectation of random errors is zero m NS deviates from the mathematical expectation by an amount less than d= 15 is equal to:

EXAMPLE 4... The machine produces balls. The ball is considered good if the deviation NS the diameter of the ball from the design size in absolute value is less than 0.7 mm. Assuming that the random variable NS distributed normally with a standard deviation of 0.4 mm, find the average number of good balls among 100 manufactured ones.

Solution. Random value NS- deviation of the ball diameter from the design size. The mathematical expectation of the deviation is zero, i.e. M(NS)=m= 0. Then the probability that the normal random variable NS deviates from the mathematical expectation by an amount less than d= 0.7, is equal to:

It follows that approximately 92 balls out of 100 will be usable.

EXAMPLE 5. Prove the rule "3 s».

Solution. The probability that a normal random variable NS deviates from the mathematical expectation by an amount less than d = 3s, is equal to:

EXAMPLE 6. Random value NS normally distributed with mathematical expectation m= 10. Hit probability NS in the interval (10, 20) is 0.3. What is the probability of hitting NS in the interval (0, 10)?

Solution. Normal curve symmetrical about a straight line NS=m= 10; therefore, the areas bounded from above by the normal curve and from below by the intervals (0, 10) and (10, 20) are equal to each other. Since the areas are numerically equal to the hit probabilities NS in the appropriate interval, then.

Local and integral Laplace theorems

This article is a natural continuation of the lesson about independent tests where we met by the Bernoulli formula and worked out typical examples on the topic. The local and integral Laplace theorems (Moivre-Laplace) solve a similar problem with the difference that they are applicable to a fairly large number of independent tests. There is no need to obscure the words "local", "integral", "theorems" - the material is mastered with the same ease with which Laplace patted Napoleon's curly head. Therefore, without any complexes and preliminary remarks, we will immediately consider a demo example:

The coin is flipped 400 times. Find the probability that heads will land 200 times.

According to the characteristic features, it should be applied here Bernoulli formula ... Let's remember the meaning of these letters:

- the probability that in independent tests a random event will occur exactly once;
– binomial coefficient;
- the probability of occurrence of an event in each test;

In relation to our task:
- the total number of tests;
- the number of throws in which heads must fall;

Thus, the probability that as a result of 400 tosses of a coin, heads will fall exactly 200 times: ... Stop, what to do next? The micro calculator (at least mine) did not cope with the 400th degree and capitulated to factorials... But I didn't want to count through the work =) Let's use standard Excel function, which managed to process the monster:.

I draw your attention to what is received precise meaning and such a solution seems to be ideal. At first sight. Here are some compelling counterarguments:

- firstly, the software may not be at hand;
- and secondly, the solution will look outside the box (with a considerable probability you will have to reconsider);

Therefore, dear readers, in the near future we are waiting for:

Local Laplace theorem

If the probability of the occurrence of a random event in each test is constant, then the probability that the event occurs exactly once in the tests is approximately equal to:
, where .

In this case, the larger, the calculated probability will better approximate the exact value obtained (at least hypothetically) by the Bernoulli formula. The recommended minimum number of tests is approximately 50-100, otherwise the result may be far from the truth. In addition, the local Laplace theorem works the better, the closer the probability to 0.5, and vice versa - it gives a significant error at values ​​close to zero or one. For this reason, another criterion for the effective use of the formula is the fulfillment of the inequality () .

So, for example, if, then the application of the Laplace theorem for 50 tests is justified. But if and, then the approximation (to exact value) will be bad.

Why and a special function we will talk in the lesson about normal probability distribution, but for now we need a formal computational side of the issue. In particular, an important fact is parity this function: .

Let's formalize the official relationship with our example:

Problem 1

The coin is flipped 400 times. Find the probability that heads will come up exactly:

a) 200 times;
b) 225 times.

Where to begin solution? First, let's write down the known values ​​so that they are before our eyes:

- the total number of independent tests;
- the probability of getting heads in each throw;
- the probability of getting tails.

a) Let us find the probability that in a series of 400 throws heads will fall out exactly once. Due to the large number of tests, we use the local Laplace theorem: , where .

At the first step, we calculate the required value of the argument:

Next, we find the corresponding value of the function:. This can be done in several ways. First of all, of course, direct calculations suggest themselves:

Rounding is carried out, as a rule, to 4 decimal places.

The disadvantage of direct calculation is that not every micro-calculator can digest the exponent, and besides, the calculations are not very pleasant and take time. Why suffer so much? Use calculator by territory (point 4) and get values ​​instantly!

In addition, there is function value table, which is in almost any book on the theory of probability, in particular, in the textbook V.E. Gmurman... Download, who has not yet downloaded - there is generally a lot of useful information ;-) And be sure to learn how to use a spreadsheet (right now!)- there may always be no suitable computer technology at hand!

At the final stage, we will apply the formula :
- the probability that with 400 tosses of a coin, heads will fall out exactly 200 times.

As you can see, the result obtained is very close to the exact value calculated by Bernoulli formula.

b) Let us find the probability that in a series of 400 tests, heads will fall out exactly once. We use the local Laplace theorem. One, two, three - and you're done:

Is the required probability.

Answer:

The next example, as many have guessed, is devoted to childbirth - and this is for you to decide on your own :)

Task 2

The probability of having a boy is 0.52. Find the probability that among 100 newborns there will be exactly: a) 40 boys, b) 50 boys, c) 30 girls.

Round off the results to 4 decimal places.

... It is interesting that the phrase "independent tests" sounds here =) By the way, the real statistical probability the birth of a boy in many regions of the world ranges from 0.51 to 0.52.

An approximate sample of task design at the end of the lesson.

Everyone noticed that the numbers turn out to be small enough, and this should not be misleading - after all, we are talking about the probabilities taken separately, local values ​​(hence the name of the theorem). And there are many such values, and, figuratively speaking, the probability "should be enough for everyone." True, many events will almost impossible.

Let me explain the above using an example with coins: in a series of four hundred tests, the eagle can theoretically fall from 0 to 400 times, and these events form full group:

However, most of these values ​​are a mere minuscule, so, for example, the probability that heads fall 250 times is already one ten million:. About values ​​like tactfully keep silent =)

On the other hand, the modest results should not be underestimated: if it is only about, then the probability that heads will fall out, say, from 220 to 250 times will be quite noticeable.

Now let's think: how to calculate this probability? Do not count by addition theorem for the probabilities of inconsistent events amount:

These values ​​are much simpler unite... And the combination of something, as you know, is called integrating:

Laplace's integral theorem

If the probability of occurrence of a random event in each test is constant, then the probability that in trials the event will come no less and no more times (from to times inclusive), is approximately equal to:

In this case, the number of tests, of course, should also be large enough and the probability is not too small / high. (tentatively), otherwise the approximation will be unimportant or bad.

The function is called Laplace function, and its values ​​are again summarized in a standard table ( find and learn to work with it !!). A microcalculator will not help here, since the integral is non-variable. But Excel has a corresponding functionality - use point 5 design layout.

In practice, the following values ​​are most common:
- rewrite it in your notebook.
Starting with, we can assume that, or, to put it more strictly:

Moreover, the Laplace function odd: , and this property is actively exploited in tasks that have already been waiting for us:

Problem 3

The probability of hitting the target by the shooter is 0.7. Find the probability that with 100 shots the target will be hit 65 to 80 times.

I chose the most realistic example, otherwise I found several problems in which the shooter makes thousands of shots =)

Solution: in this problem we are talking about repeated independent tests, and their number is quite large. According to the condition, it is required to find the probability that the target will be hit at least 65, but not more than 80 times, which means that you need to use the Laplace integral theorem:, where

For convenience, let's rewrite the initial data in a column:
- total shots;
- the minimum number of hits;
- the maximum number of hits;
- the probability of hitting the target with each shot;
- the probability of a miss with each shot.

Hence, Laplace's theorem will give a good approximation.

Let's calculate the values ​​of the arguments:

I draw your attention to the fact that the work does not have to be completely extracted from the root (as authors of problems like to "adjust" the numbers)- without a shadow of a doubt, extract the root and round the result; I'm used to leaving 4 decimal places. But the obtained values ​​are usually rounded to 2 decimal places - this tradition comes from function value tables where the arguments are presented exactly as they are.

We use the above table or calculation layout by terver (point 5).
As a written commentary, I advise you to put the following phrase: we find the values ​​of the function according to the corresponding table:

- the probability that with 100 shots the target will be hit from 65 to 80 times.

Be sure to use the odd function! Just in case, I will write in detail:

The fact is that function value table contains only positive "x", and we are working (at least according to the "legend") with a table!

Answer:

The result is most often rounded to 4 decimal places. (again according to the table format).

For an independent solution:

Problem 4

The building has 2500 lamps, the probability of each of them being switched on in the evening is 0.5. Find the probability that no less than 1250 and no more than 1275 lamps will be switched on in the evening.

A rough example of finishing at the end of the lesson.

It should be noted that the tasks under consideration are very often encountered in an "impersonal" form, for example:

Some experiment is made, in which a random event can appear with a probability of 0.5. The experiment is repeated 2500 times under unchanged conditions. Determine the probability that in 2500 experiments the event will occur from 1250 to 1275 times

And similar formulations are above the roof. Due to the stereotyped tasks, they often try to disguise the condition - this is the "only chance" to somehow diversify and complicate the solution:

Problem 5

The institute has 1000 students. The dining room has 105 seats. Each student goes to the cafeteria during the big break with a probability of 0.1. What is the likelihood that on a typical school day:

a) the dining room will be no more than two-thirds full;
b) there will not be enough seats for everyone.

I would like to draw your attention to the essential clause “on a NORMAL school day” - it ensures that the situation is relatively unchanged. After the holidays, significantly fewer students may come to the institute, and a hungry delegation may descend on the "Open Day" =) That is, on an "unusual" day, the probabilities will differ markedly.

Solution: we use the Laplace integral theorem, where

In this task:
- all students at the institute;
- the likelihood that the student will go to the cafeteria during a big break;
- the probability of the opposite event.

a) Calculate how many seats are two-thirds of the total: seats

Find the probability that on a typical school day, the cafeteria will be no more than two-thirds full. What does it mean? This means that from 0 to 70 people will come to the big break. The fact that no one will come or only a few students will come - there are events almost impossible, however, in order to apply Laplace's integral theorem, these probabilities should still be taken into account. Thus:

Let's calculate the corresponding arguments:

As a result:

- the likelihood that on a typical school day the cafeteria will be no more than two-thirds full.

Reminder : for, the Laplace function is assumed to be equal.

Swarm, however =)

b) Event "There will not be enough seats for everyone" consists in the fact that from 106 to 1000 people will come to the dining room at the big break (the main thing is to seal well =)). It is clear that the high attendance is incredible, but nevertheless: .

We calculate the arguments:

Thus, the probability that there will not be enough seats for everyone:

Answer:

Now let's dwell on one important nuance method: when we perform calculations on a single segment, then everything is "cloudless" - decide according to the considered template. However, in the case of consideration full group of events should show a certain accuracy... Let me explain this point using the example of the problem just analyzed. At the point "bh" we found the probability that there will not be enough seats for everyone. Further, according to the same scheme, we will calculate:
- the likelihood that there will be enough seats.

Since these events opposite, then the sum of the probabilities should be equal to one:

What's the matter? - everything seems to be logical here. The point is that the Laplace function is continuous, but we did not take into account interval from 105 to 106. This is where a piece of 0.0338 disappeared. That's why by the same standard formula should calculate:

Well, or even simpler:

The question arises: what if we found FIRST? Then there will be another version of the solution:

But how can this be ?! - different answers are obtained in two ways! It's simple: Laplace's integral theorem is a method approximate calculations, and therefore both are acceptable.

For more accurate calculations, you should use by the Bernoulli formula and, for example, the Excel function BINOMDIST... As a result its application we get:

And I express my gratitude to one of the site visitors who drew attention to this subtlety - it fell out of my field of vision, since the study of a complete group of events is rarely met in practice. Those interested can familiarize themselves with

Bayes' formula

Events B 1, B 2, ..., B n are incompatible and form a complete group, i.e. P (B 1) + P (B 2) + ... + P (B n) = 1. And let the event A can occur only when one of the events B 1, B 2,…, B n appears. Then the probability of event A is found by the formula for the total probability.

Let event A have already happened. Then the probabilities of hypotheses В 1, В 2, ..., В n can be re-estimated using the Bayes formula:

Bernoulli's formula

Let n independent tests be carried out, in each of which event A may or may not occur. The probability of occurrence (not occurrence) of event A is the same and is equal to p (q = 1-p).

The probability that in n independent tests event A will occur exactly the same time (according to Fig, in what sequence) is found by the Bernoulli formula:

The probability that in n independent trials the event will occur:

a). Less times P n (0) + P n (1) +… + P n (k-1).

b). More to times P n (k + 1) + P n (k + 2) +… + P n (n).

v). at least k times P n (k) + P n (k + 1) +… + P n (n).

G). at most k times P n (0) + P n (1) +… + P n (k).

Local and integral Laplace theorems.

We use these theorems when n is large enough.

Local Laplace theorem

The probability that in n independent tests the event will occur exactly `k 'times is approximately equal to:

The table of functions for positive values ​​(x) is given in Gmurman's problem book in Appendix 1, pp. 324-325.

Since it is even (), we use the same table for negative values ​​(x).

Laplace's integral theorem.

The probability that in n independent tests the event will occur at least `k 'times is approximately equal to:

Laplace function

A table of functions for positive values ​​is given in Gmurman's problem book in Appendix 2, pp. 326-327. For values ​​greater than 5, we put Ф (х) = 0.5.

Since the Laplace function is odd Ф (-х) = - Ф (х), then for negative values ​​(х) we use the same table, only the values ​​of the function are taken with a minus sign.

The law of probability distribution for a discrete random variable

Binomial distribution law.

Discrete- a random variable, the possible values ​​of which are separate isolated numbers, which this quantity takes with certain probabilities. In other words, the possible values ​​of a discrete random variable can be numbered.

The number of possible values ​​of a discrete random variable can be finite or infinite.

Discrete random variables are designated by capital letters X, and their possible values ​​are designated by small letters x1, x2, x3 ...

For example.

X is the number of points dropped on the dice; X takes six possible values: x1 = 1, x2 = 1, x3 = 3, x4 = 4, x5 = 5, x6 = 6 with probabilities p1 = 1/6, p2 = 1/6, p3 = 1/6 ... p6 = 1/6.

The law of distribution of a discrete random variable is called a list of its possible values ​​and the corresponding probabilities.

The distribution law can be set:

1. in the form of a table.

2. Analytically - in the form of a formula.

3. graphically. In this case, points M1 (x1, p1), M2 (x2, p2), ... Mn (xn, pn) are constructed in a rectangular coordinate system XOP. These points are connected by line segments. The resulting figure is called distribution polygon.

To write the law of distribution of a discrete random variable (x), it is necessary to list all its possible values ​​and find the corresponding probabilities.

If the corresponding probabilities are found by the Bernoulli formula, then such a distribution law is called binomial.

Example No. 168, 167, 171, 123, 173, 174, 175.

Numerical values ​​of discrete random variables.

Mathematical expectation, variance and standard deviation.

The characteristic of the average value of a discrete random variable is the mathematical expectation.

Mathematical expectation a discrete random variable is the sum of the products of all its possible values ​​by their probabilities. Those. if the distribution law is given, then the mathematical expectation

If the number of possible values ​​of a discrete random variable is infinite, then

Moreover, the series on the right side of the equality converges absolutely, and the sum of all probabilities pi is equal to one.

Mathematical expectation properties.

1.M (C) = C, C = post.

2.M (Cx) = CM (x)

3.M (x1 + x2 + ... + xn) = M (x1) + M (x2) + ... + M (xn)

4. M (x1 * x2 * ... * xn) = M (x1) * M (x2) * ... * M (xn).

5. For the binomial distribution law, the mathematical expectation is found by the formula:

The characteristic of the scattering of possible values ​​of a random quantity around the mathematical expectation is the variance and the standard deviation.

Dispersion a discrete random variable (x) is called the mathematical expectation of the square of the deviation. D (x) = M (x-M (x)) 2.

It is convenient to calculate the variance by the formula: D (x) = M (x 2) - (M (x)) 2.

Dispersion properties.

1.D (S) = 0, S = constant.

2.D (Cx) = C 2 D (x)

3. D (x1 + x2 + ... + xn) = D (x1) + D (x2) + ... + D (xn)

4. Dispersion of the binomial distribution law

Mean square deviation a random variable is called the square root of the variance.

examples. 191, 193, 194, 209, d / z.

Cumulative distribution function (IGF, FD) of probabilities of a continuous random variable (NSV). Continuous- a quantity that can take all values ​​from a certain finite or infinite interval. The number of possible values ​​of the NWS is available and cannot be renumbered.

For example.

The distance that the projectile travels when fired is NSV.

IGF is called the function F (x), which determines, for each value of x, the probability that the NSV X will take the value X<х, т.е. F(x)=Р(X

Often, instead of an IGF, they say FR.

Geometrically, the equality F (x) = P (X

IF properties.

1. The IF value belongs to the interval, i.e. F (x).

2. IF is a non-decreasing function, i.e. x2> x1 ,.

Corollary 1. The probability that the NSV X will take the value included in the interval (a; c) is equal to the increment of the integral function on this interval, i.e.

P (a

Corollary 2. The probability that NSV X will take one definite value, for example, x1 = 0, is equal to 0, i.e. P (x = x1) = 0.

3. If all possible values ​​of the NSV X belong to (a; c), then F (x) = 0 for x<а, и F(x)=1 при х>v.

Corollary 3. The following limit relations are valid.

Differential distribution function (DFD) of the probabilities of a continuous random variable (DCV) (probability density).

DF f (x) the probability distribution of the NSV call the first derivative of IGF:

Probability density (PV) is often said instead of FDR.

It follows from the definition that, knowing the IF F (x), one can find the DF f (x). But the reverse transformation is also performed: knowing the DF f (x), one can find the IF F (x).

The probability that NSV X will take a value belonging to (a; c) is found:

A). If IF is given - Corollary 1.

B). If DF is given

DF properties.

1. DF - not negative, i.e. ...

2. the improper integral of the DF within () is equal to 1, i.e. ...

Corollary 1. If all possible values ​​of the NSV X belong to (a; c), then.

Examples. No. 263, 265, 266, 268, 1111, 272, d / z.

Numerical characteristics of NSV.

1. The mathematical expectation (MO) of NSV X, the possible values ​​of which belong to the entire OX axis, is determined by the formula:

If all possible values ​​of NSV X belong to (a; c), then MO is determined by the formula:

All MO properties indicated for discrete quantities are also preserved for continuous quantities.

2. Dispersion of NSV X, the possible values ​​of which belong to the entire OX axis, is determined by the formula:

If all possible values ​​of NSV X belong to (a; c), then the variance is determined by the formula:

All dispersion properties specified for discrete quantities are retained for continuous quantities.

3. The mean square deviation of NSV X is determined in the same way as for discrete values:

Examples. No. 276, 279, X, d / z.

Operating calculus (OI).

OI is a method that allows you to reduce the operations of differentiation and integration of functions to simpler actions: multiplication and division by the argument of the so-called images of these functions.

The use of OI facilitates the solution of many problems. In particular, problems of integrating LDEs with constant coefficients and systems of such equations, reducing them to linear algebraic ones.

Originals and images. Laplace transforms.

f (t) -original; F (p) -image.

The transition f (t) F (p) is called Laplace transform.

The Laplace transform of a function f (t) is called F (p), which depends on a complex variable and is defined by the formula:

This integral is called the Laplace integral. For the convergence of this improper integral, it suffices to assume that in the interval f (t) it is piecewise continuous and for some constants M> 0 and satisfies the inequality

A function f (t) with such properties is called original, and the transition from the original to its image is called Laplace transform.

Laplace transform properties.

Direct determination of images by formula (2) is usually difficult and can be greatly facilitated by using the properties of the Laplace transform.

Let F (p) and G (p) be the images of the originals f (t) and g (t), respectively. Then the following properties-relations hold:

1.C * f (t) C * F (p), C = const - the property of homogeneity.

2.f (t) + g (t) F (p) + G (p) -property of additivity.

3. f (t) F (p-) -displacement theorem.

transition of the n-th derivative of the original into the image (the theorem of differentiation of the original).