The concept of harmonic functions.

Harmonic function- real function $U$, defined and twice continuously differentiable on Euclidean space $D$(or an open subset thereof) satisfying Laplace's equation:

$\backslash Delta\; U\; =\; 0.\backslash$

Where $\backslash Delta=\backslash sum\_(i=1)^n\backslash frac(\backslash partial^2)(\backslash partial\; x\_i^2)$- Laplace operator, that is, the sum of second derivatives over all rectangular Cartesian coordinates x i (n= dim D- dimension of space).

For example, the harmonic function is the electrostatic potential at points where there is no charge.

Properties

Maximum principle

Function U, harmonic in the region $D$, reaches its maximum and minimum only at the boundary $\backslash partial\; D$. Thus, a harmonic function cannot have a local extremum region at an internal point, except in the trivial case of a constant at $D$ functions. However, the function may be undefined on the boundary, so it is more correct to say

Liouville's theorem

Harmonic function defined on $\backslash Bbb(R)^n$ and limited above or below, constant.

Property of the average

If the function $u$ harmonious in some ball $B(x\_0)$ centered at a point $x\_0$, then its value at the point $x\_0$ equal to its average value along the boundary of this ball or over the ball:

$u(x\_0)\; =\; \backslash frac(1)(\backslash mu(\backslash partial\; B))\; \backslash int\backslash limits\_(\backslash partial\; B)\; u\; dS\; =\; \backslash frac(1)(\backslash mu\; (B))\; \backslash int\backslash limits\_(B)\; u\; dV$

Where $\backslash mu\; (B)$- volume of the ball $B(x\_0)$ And $\backslash mu(\backslash partial\; B)$- area of ​​its border.

Conversely, any continuous function that has the property of an average for all balls lying in a certain region is harmonic in this region.

Differentiability

A function that is harmonic in a domain is infinitely differentiable in it.

Harnack's inequality

If the function $U(M)=U(x\_1,...x\_k)$, harmonic in the k-dimensional ball $Q\_r$ radius $R$ with center at some point $M\_0$, is non-negative in this ball, then for its values ​​at points $M$ inside the ball under consideration the following inequalities are valid: $((R^(k-2))(\backslash frac(R-r)((R+r)^(k-1)))U(M\_0))\backslash le(U(M))\backslash le(R^(k\; -2)\backslash frac(R+r)((R-r)^(k-1))U(M\_0))$, Where

Harnack's theorem

Let $v\_n(z)$- positive harmonic functions in some area $D$. If the row $\backslash sum\_(1)^\backslash infty\; v\_(n)(z)$ converges at least at one point in the region $D$, then it converges uniformly inside $D$.

see also

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Notes

Literature

• Vladimirov V. S., Zharinov V. V. Equations of mathematical physics. - Fizmatlit, 2004. - ISBN 5-9221-0310-X.

Excerpt characterizing the Harmonic function

“Fr... fr...” Prince Nikolai Andreich snorted.
- The prince, on behalf of his pupil... son, makes a proposition to you. Do you want or not to be the wife of Prince Anatoly Kuragin? You say yes or no! - he shouted, - and then I reserve the right to say my opinion. Yes, my opinion and only my opinion,” added Prince Nikolai Andreich, turning to Prince Vasily and responding to his pleading expression. - Yes or no?
– My desire, mon pere, is never to leave you, never to separate my life from yours. “I don’t want to get married,” she said decisively, looking with her beautiful eyes at Prince Vasily and her father.
- Nonsense, nonsense! Nonsense, nonsense, nonsense! - Prince Nikolai Andreich shouted, frowning, took his daughter by the hand, bent her to him and did not kiss her, but only bending his forehead to her forehead, he touched her and squeezed the hand he was holding so much that she winced and screamed.
Prince Vasily stood up.
– Ma chere, je vous dirai, que c"est un moment que je n"oublrai jamais, jamais; mais, ma bonne, est ce que vous ne nous donnerez pas un peu d"esperance de toucher ce coeur si bon, si genereux. Dites, que peut etre... L"avenir est si grand. Dites: peut etre. [My dear, I will tell you that I will never forget this moment, but, my dearest, give us at least a small hope of being able to touch this heart, so kind and generous. Say: maybe... The future is so great. Say: maybe.]
- Prince, what I said is everything that is in my heart. I thank you for the honor, but I will never be your son's wife.
- Well, it’s over, my dear. Very glad to see you, very glad to see you. “Come to yourself, princess, come,” said the old prince. “I’m very, very glad to see you,” he repeated, hugging Prince Vasily.
“My calling is different,” Princess Marya thought to herself, my calling is to be happy with another happiness, the happiness of love and self-sacrifice. And no matter what it costs me, I will make poor Ame happy. She loves him so passionately. She repents so passionately. I will do everything to arrange her marriage with him. If he is not rich, I will give her money, I will ask my father, I will ask Andrey. I will be so happy when she becomes his wife. She is so unhappy, a stranger, lonely, without help! And my God, how passionately she loves, if she could forget herself like that. Maybe I would have done the same!...” thought Princess Marya.

For a long time the Rostovs had no news of Nikolushka; Only in the middle of winter was a letter given to the count, at the address of which he recognized his son’s hand. Having received the letter, the count, frightened and hasty, trying not to be noticed, ran on tiptoe into his office, locked himself and began to read. Anna Mikhailovna, having learned (as she knew everything that was happening in the house) about the receipt of the letter, quietly walked into the count’s room and found him with the letter in his hands, sobbing and laughing together. Anna Mikhailovna, despite the improvement in her affairs, continued to live with the Rostovs.

Connection with analytical functions. Analytic functions are closely related to harmonic functions of two variables, i.e., to solutions of the two-dimensional Laplace equation

Indeed, differentiating the first of the analyticity conditions

We will find that the function u, the real part of the analytic function, is a harmonic function. In the same way, it is proved that the imaginary part of the analytic function is a harmonic function.

will be analytic in D. Indeed, by virtue of equation (1)

in a simply connected domain is the exact differential of some function v, which is the desired one. Thus, the conjugate harmonic functions are found by simple integration.

From the properties of analytic functions one can derive the corresponding properties of harmonic functions (if desired, one can do the opposite). Thus, we can assert that every harmonic function is infinitely differentiable. From formula (19) of the previous paragraph, by separating the real parts, we obtain the mean value theorem for harmonic functions:

belongs to the area of ​​harmony and.

This theorem is one of the fundamental facts of the theory of harmonic functions. From it, in particular, an important principle of extremum is obtained: a function that is not constant and harmonic in the domain D cannot reach either a maximum or a minimum inside D.

should reach the maximum

everywhere in D.

A natural problem arises of reconstructing a function harmonic in a domain from its boundary values. This problem is fundamental in the theory of harmonic functions and its applications and is called the Dirichlet problem. Here's how it's formulated:

The above reasoning shows that the Dirichlet problem cannot have two different solutions, i.e. it proves the uniqueness of the solution to this problem. A more subtle and difficult to prove fact is the existence of a solution to the Dirichlet problem. However, for a number of the simplest domains the existence of a solution can be proved by a direct construction.

apply the Cauchy integral formula:

and, therefore, we use the theorem

Now we subtract this equality from the previous one, having previously calculated that

1 we have

we will get

now there is a real multiplier. Separating the real parts in the last formula, we obtain the so-called Poisson integral

defines a function u(z) harmonic in a circle with given boundary values.

By transforming the Cauchy formula (4), similar to the one described, one can also obtain the Schwarz integral, which restores the function f(z) analytic in the unit disk from the boundary values ​​of its real part:

this problem can obviously be solved up to an imaginary constant.

denote, respectively, the values ​​of the real part of f at the lower and upper boundaries of the strip.

is harmonious in A. The theorem is proven by direct calculation, according to which the Laplace operator

In particular, harmonicity is preserved under conformal mappings, which are one-to-one analytical transformations.

The connection between the theory of harmonic functions and the theory of conformal mappings is also manifested in the connection between the corresponding boundary value problems. The main boundary value problem of the theory of conformal mappings is the following Riemann problem:

Implementing a conformal mapping of one of these regions to another.

It remains to return to the variable r and use the conservation of harmonicity under conformal mappings; we get the desired solution:

which is what you are looking for

the mapping f takes w = 0 to the center of the circle (Fig. 19).

In it, the function f must have a zero, and, moreover, of the first order, because in the neighborhood of zeros of higher order the analytic function is not one-to-one (it has the character of a degree there). Therefore, in a neighborhood of z0 the function f must have a Taylor expansion of the form

But then the logarithm of this

function is analytic in D, which means its real part, i.e., the function

must be harmonic in D.

Now it is not difficult to understand the intention of the constructions carried out: after all, if f maps D onto the unit disk, then it must be equal to 1 on the boundary of D, which means that, without yet knowing the conformal mapping itself, we know the boundary values ​​of function (9), they are equal

and are determined by the geometric shape of the region’s boundary and the selected point z0 (Fig. 19). To find the desired conformal mapping, it is necessary, therefore, to perform the following operations: 1) using known boundary values ​​(1СН, construct a harmonic in D

function u(z) (Dirichlet problem), 2) find the function v(z) harmonically conjugate to u (integration). Now we know the function

where the required mapping is found using the formula

I would like to draw attention to some subtleties associated with the constructed solution. From the construction it is clear that the function f is analytic in D and that on the boundary of D its modulus is equal to 1. However, it remains to be proven that this function is a one-to-one mapping of D onto the unit disk. This can be done by direct (but by no means simple) verification. If we are confident that our problem is solvable (that is, we know how to prove the theorem for the existence of a conformal mapping D onto a circle), then such a check is unnecessary - the above reasoning shows that if the required mapping exists, then it can certainly be restored using the formula (eleven).

will be the desired conformal mapping.

Consider a function U that is harmonic in a bounded domain (D) with a surface (S). Assuming that U is continuous along with second-order derivatives up to (S) and applying Green’s formula (6) to this function U and to the harmonic function, we obtain, by virtue of

i.e., we have the first property of the harmonic function: the integral of the normal derivative of the harmonic function over the surface of the region is equal to zero.

If we apply formula (9) to the harmonic function U, then, due to , we obtain

This gives us the second property of the harmonic function: the value of the harmonic function at any point inside the region is expressed through the values ​​of this function and its normal derivative on the surface of the region by formula (13).

Note that the integrals in formulas (12) and (13) do not contain second-order derivatives of the function and for the applicability of these formulas it is enough to assume that the harmonic function is continuous along with the first-order derivatives up to (S). To verify this, it is enough to slightly compress the surface (S), write formulas (12) and (13) for the compressed region (D), in which there is continuity and second-order derivatives up to the surface, and then go to the limit, expanding (D ) to (D). Compression can be done, for example, by plotting the same small segment of length 8 on the internal normal to (S) at each of its points. The ends of these segments form a new (compressed) surface. In this case, the surface (S) must be such that the described transformation for all sufficiently small 8 leads to a surface that does not intersect itself and is piecewise smooth. This issue will be discussed in detail in Volume IV.

Let us apply formula (13) to a special case of a region, namely to a sphere with center at and radius R, assuming, of course, that

the function U is harmonic in this sphere and is continuous with first-order derivatives up to its surface (21)

In this case, the direction of the outer normal coincides with the direction of the radius of the sphere, so we will have

and formula (13) gives

But on the surface of a sphere the quantity has a constant value R, so

or, by virtue of (12), we will finally have

This formula expresses the third property of the harmonic function: the value of the harmonic function at the center of the sphere is equal to the arithmetic mean value of this function on the surface of the sphere, that is, equal to the integral of the values ​​of the function over the surface of the sphere, divided by the area of ​​this surface.

From this property the following fourth property of the harmonic function follows almost obviously:

A function that is harmonic within a region and continuous up to the boundary of the region reaches its maximum and minimum values ​​only at the boundary of the region, except in the case when this function is constant. Let us give a detailed proof of this statement. Let it reach its greatest value at some internal point of the region where the harmonic function is. Let us construct a sphere with center and radius belonging to formula (14) and replace the integrand function U with its largest value on the sphere. Thus we obtain

moreover, the equal sign occurs only in the case when U on the sphere is a constant equal to . Since, by assumption, and is the greatest value in, we can assert that the equal sign holds, and that, therefore,

Equal to the constant inside and on the surface of any sphere with the center belonging to D. Let us show that it follows from this that there is a constant in the entire area

Let N be any point lying inside D. We need to show that We connect to N with a line of finite length, for example a broken line, lying inside and let d be the shortest distance from the boundary S of the region D (d is a positive number). By virtue of what was proved above, it is equal to a constant in a ball with center and radius d. Let be the last point of intersection of the line with the surface of the mentioned ball, if we count from We have and, according to what was proven above, is equal to a constant in a ball with center and radius d. Let the last point of intersection of l with the surface of this ball. As above, the function is equal to a constant in a ball with center and radius d, etc. By constructing a finite number of such balls, we will be convinced that this is what we needed to prove. It can also be shown that D cannot have either maxima or minima inside. Using the proven property of harmonic functions, it is very easy to show that the internal Dirichlet problem, which we mentioned in, can have only one solution. Indeed, if we assume that there are two functions that are harmonic inside D and take the same limit values ​​on the surface S of this region, then the difference will also satisfy the Laplace equation inside D, i.e., it will be a harmonic function, and its limit values ​​on the surface 5 equal to zero everywhere. From here, in view of what was proved above, it immediately follows that it vanishes identically throughout the entire region, for otherwise it would have to reach a positive maximum value or a negative minimum value inside, which is impossible. Thus, the two solutions to the Dirichlet problem must coincide throughout the entire domain D. The uniqueness of the external Dirichlet problem is proved in exactly the same way, taking into account that, by condition, at an infinitely distant point the harmonic function must vanish.

Completely similar properties are obtained for harmonic functions on the plane. In this case, instead of formula (13) we will have the formula

and the mean value theorem will be expressed as

where is a circle with center and radius R. For the external Dirichlet problem at an infinitely distant point, it is not necessary to vanish, as in the three-dimensional case, but only the existence of some finite limit, and the uniqueness of the Dirichlet problem must be proven differently than in the previous case. We will present this proof in Volume IV, where we will consider the Dirichlet and Neumann problems in more detail.

Let us note now that any constant is a harmonic function that satisfies the limit condition

from which it is clear that if an arbitrary constant is added to the solution of the Neumann problem, then the resulting sum will also be a solution to the Neumann problem with the same limit values, i.e., the solution to the Neumann problem is determined up to an arbitrary constant term. From formula (12) it also follows that the function included in the limit condition of the internal Neumann problem cannot be arbitrary, but must satisfy the condition

In conclusion, we also note that formula (13) is also valid in the case when there is a harmonic function in an infinite region formed by a part of space located outside the surface S. In this case, one only needs to make an assumption about the order of smallness at infinity, i.e., at infinite distance of the point M. It is sufficient (and necessary) to assume that with infinite distance the inequalities take place

The function a(t) is called harmonic if it varies according to a sinusoidal or cosine law:

a(t) = A m Cos(ωt + φ) = A m sin(ωt + ψ).

Here the argument υ(t) = ωt + ψ is called the phase. The value ψ = φ + π/2, equal to the phase value at t = 0, is called the initial phase. The largest value of the function is the amplitude A m, the smallest value is (–A m).

The phase of the harmonic function increases linearly with time. The rate ω of its change is called the angular frequency and is measured in
rad/s. The harmonic function is the simplest type of periodic function. The value f, the reciprocal of the period of the function T, is called the linear frequency and is measured in hertz, denoted Hz.

The steady state of a circuit is one in which the law of change in voltage and current does not change during the entire period under study.

time gap. Otherwise, the regime is transitional.

Let us consider established processes.

Let's plot the harmonic function:

1. Select a scale. Along the abscissa axis is the phase ωt to determine the period of the 2π function. The y-axis is amperes (if it is a function of current) or volts (if it is a function of voltage). Let us set aside the amplitude of the function A m (Fig. 2.2):

3. Shift the constructed function along the abscissa axis by the value of the initial phase ψ. If ψ > 0, then we shift it to the left, that is, the function a(ωt) advances the origin along the abscissa by the amount ψ. If ψ< 0, то сдвигаем вправо, то есть

the function a(ωt) lags behind the origin by the amount ψ.

For example, with , we get (Fig. 2.4):

Example 9. Plot a graph of the current function i(t) = 2 Sin(ωt + ) A.

1. Select the scale along the ordinate axes (Fig. 2.5).

2. Construct the function i´(t) = 2 Sin(ωt +0) A (Fig. 2.6).

3. Shift the constructed function along the x-axis to the left, since

Ψ = > 0 (Fig. 2.7):

3. We shift the constructed function along the x-axis to the right, since

Ψ = – < 0 (рис. 2.10).

Average and effective values ​​of harmonic currents and

Voltage

The average value of the periodic function i(t), u(t), over the period T is determined by the expression:

The average value does not depend on the time t 0 .

The average value over the period of the harmonic function (which is current and voltage (emf)) is zero.

The effective value of a periodic function i(t), u(t), is the mean square value of this function over the period T:

.

In its physical meaning, the effective value of a periodic current per period is a direct current that, passing through a constant resistance, emits the same amount of heat as the given current.

Effective value I, U, E of the harmonic function i(t) , u(t), in

Times less than the amplitude

) - 1 , ) - 1 , ) - 1 .

Hence,

Example 11. Current i(t) = 5Sin(ωt + ). Determine the average, effective and
its amplitude value.

The average value of I CP = 0, since i(t) is a harmonic function. Amplitude value I m = 5 A, and effective value ) - 1 = 0.707 5 = 3.535 A.

Operations with complex numbers

In mathematics and electrical engineering, the imaginary unit, which is the basis of complex numbers, is widely used.

In general, complex quantities, with the exception of current and voltage, are indicated by a symbol and a thick line under it: A. Complex numbers have five forms of representation.

Algebraic

A= a + jb; a = Re [ A]; b = Im[ A] .

here a and b are the real and imaginary components of the number, respectively A .

Indicative

A= A e j ψ ,

where A = − modulus of number A, − number argument A.

Polar

Trigonometric

A= АCosψ + jАSinψ,

where ACosψ = ​​a; ASinψ = b.

Geometric – number as a vector on the complex plane (Fig. 2.11).

Two complex numbers are called conjugate if their real components coincide and their imaginary components differ only in signs. Conjugate of a number A complex number is denoted. If A= a + jb, then = a – jb.

Addition and subtraction of complex numbers can be done in algebraic and geometric forms, but in calculations - only in algebraic form:

A 1 + A 2 = (a 1 + jb 1) ± (a 2 + jb 2) = (a 1 ± a 2) + j(b 1 ± b 2)

Multiplication and division are best done in exponential form.

A 1 · A 2 = A 1 · A 2 · = A 1 · A 2 · ,

=

Example 12. Given A 1 = 2 + j3; A 2 = 5 – j10. Determine the sum and difference
numbers A 1 and A 2 .

A 1 + A 2 = 2 + j3 + (5 – j10) = 7 – j7;

A 1 – A 2 = 2 + j3 – (5 – j10) = – 3 + j13.

Example 13. Given A 1 = 10 e j 30° ; A 2 = 20 e –j6 0° . Define product
and quotient of numbers A 1 and A 2 .

Ā 3 = A 1 · A 2 = 10 е j 30° 20 е –j6 0° = 200 е –j3 0° .

Ā 4 = A 1 = 10e j 30º (20e –j6 0°) –1 = 0.5e j 90º = 0.5j.

Very often in calculations it becomes necessary to switch from the exponential form of a complex number to the algebraic one or vice versa. Transition algorithms are proposed.

Algorithm transition from the exponential A·e jψ form to the algebraic a + jb.

1. Determine Cosψ.

2. Determine А·Cosψ = ​​а (reset).

3. Define Sin ψ.

4. Determine А·Sin ψ = b.

Algorithm transition from the algebraic a + jb form to the exponential A·e jψ.

1. Define – calculated argument ψ.

2. The true argument ψ is determined by ψ CALC depending on the quadrant in accordance with the diagram (Fig. 2.12):

3. Determine Sin ψ CALC.

4. Define .

Example 14. Translate A= 10· in algebraic form.

A= 10· ; .

10·0.865 + j10·0.5 =8.65 + j5.

Translate A=3 + j6 in exponential form.

; ψ RASCH = arctan 2 = 63°; A = 6.7;

A= 6.7e j63° .

2.4. Representation of a harmonic function on a complex one
plane

The steady-state values ​​of currents and voltages of linear circuits under the influence of harmonic signals can, in principle, be found by drawing up and solving differential equations corresponding to these processes. However, this is a rather difficult path.

At the end of the 19th century, American engineers A. Kennelly and I. Steinmetz proposed a simpler way, based on the representation of harmonic functions of time in the form of complex numbers, that is, on the transfer of the original functions from the time domain to the frequency domain.

Let us introduce the concept of complex amplitude values ​​of harmonic functions of current (voltage, emf). To do this, let us represent each of these functions as a vector on the complex plane, the length of which is equal to the amplitude A m. At the same time, it rotates with a circular frequency ω counterclockwise (Fig. 2.13).

+1

If you stop a vector at an arbitrary time t, then its projection a(t) onto the imaginary axis will be determined:

а(t) = А m·Sin (ωt + ψ) .

At t = 0 a(0) = A m ·Sinψ.

Thus, the harmonic function а(t) = А m · Sin (ωt + ψ) corresponds to a complex number A m = A m e jψ .

Similar to harmonic influences

i(t) = I m Sin (ωt + ψ i), u(t) = U m Sin (ωt + ψ u) and e(t) = E m Sin (ωt + ψ e) current value or The voltage of the harmonic function is a complex number, the modulus of which is equal to the effective value of the current or voltage, and the argument is equal to the initial phase of the harmonic function. = 10 ()‾ 1 = 7.07 V.

The reverse transformation is also valid.

The complex effective value of the current is known = 0.2e j 70° A at a frequency ω = 100 rad/s. Find the harmonic function of the current.

i(t) = I m Sin (ωt+ψ i) = I Sin (ωt+ψ i) = 0.2 Sin (100t+70°) =

HARMONIC FUNCTION

- function continuous with its second derivatives in the domain G and satisfying in Laplace's G equation=0. G. f. arise when solving problems of electrostatics, the theory of gravitation, hydrodynamics of an incompressible fluid, the theory of elasticity, etc. G. f. are, for example, the potentials of forces at points outside the sources of their field, the potential of velocities of an incompressible fluid. The simplest example of G. f. serves the foundation. solution of the Laplace equation describing the potential of a point source. Any G. f. can be represented as the sum of the potentials of a simple and double layer, expressed through the values ​​of the G. f. And and its normal derivative: if r- distance from any point P0 inside G to variable point P on the border S, then in the case of three dimensions

For G. f. the extremum principle is valid: a function that is internally harmonic G and continuous in a closed region G+S, reaches its maximum and minimum values ​​only at S, except for the case when this function is constant. This principle allows us to establish the general properties of physical quantities without resorting to calculations. For example, in electrostatics, Earnshaw’s theorem follows from it. A convenient method for solving problems for geometric functions. on the plane is given by the theory of functions of a complex variable z=x+iy. If w=u+iv - analytical function from z to G, That u(x, y)And v(x, y) are G. f. V G. Therefore, plural problems can be solved using a conformal mapping of the domain G into a certain standard area (circle, half-plane). Boundary conditions for G. f. determine the corresponding boundary value problems, of which the first boundary value problem is more common, or Dirichlet problem, when at the border S G. f. takes the given values, and the second boundary value problem, or Neumann problem, when at each point S the normal derivative of the G. function is given.

Lita.: Smirnov V.I., Course of Higher Mathematics, vol. 2, 21st ed., M., 1974; Sobolev S.L., Equations of Mathematical Physics, 4th ed., M., 1966.

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Function Function (mat.). – A few more comments should be added to what has been said. Let us assume that y is a function of the independent variable x. It may happen that this function is not defined for all values ​​of x, but only for some. For example, Ф.у = 1. 2. 3:.. (x – 1).x is defined only for integers

Harmonic proportion

From the book Great Soviet Encyclopedia (GA) by the author TSB

FUNCTION

From the book The Newest Philosophical Dictionary author Gritsanov Alexander Alekseevich

SUM function

From the book Database Processing in Visual Basic®.NET author McManus Geoffrey P

SUM Function Your ability to summarize results is not limited to simply counting records. Using the SUM function, you can generate total results for all returned records for any numeric field. For example, to create a query that generates totals for

uni() function

From the book Fiction Book Designer 3.2. Quick Guide by Izekbis

uni() function

From Fiction Book Designer Quick Guide author author unknown

uni() function Finding/replacing a character by its Unicode number can also be done using the uni() function. Example of the uni() function: Boouni(107,32)Designer will find the word Book

Function not

From the book XSLT Technology author Valikov Alexey Nikolaevich

sum function

From the book XSLT Technology author Valikov Alexey Nikolaevich

Part 1. The complete function of management in crowd-“elitism” and in real democracy 1.1. The complete management function and the primitive practice of its implementation in the life of society

From the book “About the Current Moment” No. 7(79), 2008. author USSR Internal Predictor

Part 1. The complete function of management in crowd-“elitism” and in real democracy 1.1. The complete management function and the primitive practice of its implementation in the life of society In a fairly general theory of management (DOTU) there is the concept of “complete management function”. Full function