An irrational function of a variable is a function that is formed from a variable and arbitrary constants using a finite number of operations of addition, subtraction, multiplication (raising to an integer power), division and extraction of roots. An irrational function differs from a rational function in that the irrational function contains operations for extracting roots.
There are three main types irrational functions, the indefinite integrals of which are reduced to integrals of rational functions. These are integrals containing roots of arbitrary integer degrees from a linear-fractional function (roots can be of different degrees, but from the same linear-fractional function); integrals of differential binomial and integrals with square root of a square trinomial.
Important note. The roots are ambiguous!
When calculating integrals containing roots, expressions of the form are often encountered, where is some function of the variable of integration. It should be borne in mind that. That is, for t> 0, | t | = t... At t< 0, | t | = - t. Therefore, when calculating such integrals, it is necessary to consider separately the cases t> 0 and t< 0 ... This can be done by writing signs or where necessary. Assuming that the upper sign refers to the case t> 0 , and the lower one - to the case t< 0 ... Upon further transformation, these signs, as a rule, cancel each other out.
The second approach is also possible, in which the integrand and the result of integration can be considered as complex functions on complex variables. Then you can not follow the signs in radical expressions. This approach is applicable if the integrand is analytical, that is, a differentiable function of a complex variable. In this case, both the integrand and the integral of it are multivalued functions. Therefore, after integration, when substituting numerical values, it is necessary to select a single-valued branch (Riemann surface) of the integrand, and for it choose the corresponding branch of the integration result.
Fractional Linear Irrationality
These are integrals with roots of the same linear fractional function:
,
where R is a rational function, are rational numbers, m 1, n 1, ..., m s, n s are integers, α, β, γ, δ are real numbers.
Such integrals are reduced to an integral of a rational function by substitution:
, where n is the common denominator of the numbers r 1, ..., r s.
The roots may not necessarily be from a linear fractional function, but also from a linear one (γ = 0, δ = 1), or on the variable of integration x (α = 1, β = 0, γ = 0, δ = 1).
Here are examples of such integrals:
,
.
Integrals of differential binomials
Integrals of differential binomials are:
,
where m, n, p are rational numbers, a, b are real numbers.
Such integrals reduce to integrals of rational functions in three cases.
1) If p is an integer. Substitution x = t N, where N is the common denominator of the fractions m and n.
2) If - whole. Substitution a x n + b = t M, where M is the denominator of p.
3) If - whole. Substitution a + b x - n = t M, where M is the denominator of p.
In other cases, such integrals are not expressed in terms of elementary functions.
Sometimes such integrals can be simplified using reduction formulas:
;
.
Integrals containing the square root of a square trinomial
Such integrals are of the form:
,
where R is a rational function. There are several solution methods for each such integral.
1)
With the help of transformations, lead to simpler integrals.
2)
Apply trigonometric or hyperbolic substitutions.
3)
Apply Euler substitutions.
Let's take a closer look at these methods.
1) Transformation of the integrand
Applying the formula, and performing algebraic transformations, we bring the integrand to the form:
,
where φ (x), ω (x) are rational functions.
Type I
Integral of the form:
,
where P n (x) is a polynomial of degree n.
Such integrals are found by the method of undefined coefficients using the identity:
.
Differentiating this equation and equating the left and right sides, we find the coefficients A i.
II type
Integral of the form:
,
where P m (x) is a polynomial of degree m.
Substitution t = (x - α) -1 this integral is reduced to the previous type. If m ≥ n, then the whole part of the fraction should be selected.
III type
Here we do the substitution:
.
Then the integral will take the form:
.
Further, the constants α, β must be chosen such that the coefficients at t in the denominator vanish:
B = 0, B 1 = 0.
Then the integral decomposes into the sum of integrals of two types:
,
,
which are integrated by substitutions:
u 2 = A 1 t 2 + C 1,
v 2 = A 1 + C 1 t -2.
2) Trigonometric and hyperbolic substitutions
For integrals of the form, a > 0
,
we have three main substitutions:
;
;
;
For integrals, a > 0
,
we have the following substitutions:
;
;
;
And, finally, for integrals, a > 0
,
the substitutions are as follows:
;
;
;
3) Euler substitutions
Also, integrals can be reduced to integrals of rational functions of one of three Euler substitutions:
, for a> 0;
, for c> 0;
, where x 1 is the root of the equation a x 2 + b x + c = 0. If this equation has real roots.
Elliptic integrals
In conclusion, consider integrals of the form:
,
where R is a rational function,. Such integrals are called elliptic. In general, they are not expressed in terms of elementary functions. However, there are cases when there are relations between the coefficients A, B, C, D, E in which such integrals are expressed in terms of elementary functions.
Below is an example related to return polynomials. The calculation of such integrals is performed using substitutions:
.
Example
Calculate the integral:
.
Solution
We make a substitution.
.
Here, for x> 0
(u> 0
) we take the upper sign ′ + ′. For x< 0
(u< 0
) - lower ′ - ′.
.
Answer
References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, "Lan", 2003.
A function F (x) differentiable in a given interval X is called antiderivative for function f (x), or an integral of f (x), if for any x ∈X the following equality holds:
F "(x) = f (x). (8.1)
Finding all antiderivatives for a given function is called its integration. Indefinite integral of a function f (x) on a given interval X is the set of all antiderivatives for the function f (x); designation -
If F (x) is some primitive for the function f (x), then ∫ f (x) dx = F (x) + C, (8.2)
where C is an arbitrary constant.
Integral table
Directly from the definition, we obtain the basic properties of not definite integral and a list of table integrals:
1) d∫f (x) dx = f (x)
2) ∫df (x) = f (x) + C
3) ∫af (x) dx = a∫f (x) dx (a = const)
4) ∫ (f (x) + g (x)) dx = ∫f (x) dx + ∫g (x) dx
List of table integrals
1.∫x m dx = x m + 1 / (m + 1) + C; (m ≠ -1)
3.∫a x dx = a x / ln a + C (a> 0, a ≠ 1)
4.∫e x dx = e x + C
5.∫sin x dx = cosx + C
6.∫cos x dx = - sin x + C
7. = arctan x + C
8. = arcsin x + C
10. = - ctg x + C
Variable replacement
To integrate many functions, use the method of changing the variable or substitutions, allowing to reduce integrals to tabular form.
If the function f (z) is continuous on [α, β], the function z = g (x) has a continuous derivative and α ≤ g (x) ≤ β, then
∫ f (g (x)) g "(x) dx = ∫f (z) dz, (8.3)
moreover, after integration, the substitution z = g (x) should be made on the right-hand side.
For the proof, it is enough to write the original integral in the form:
∫ f (g (x)) g "(x) dx = ∫ f (g (x)) dg (x).
For example:
Integration by parts
Let u = f (x) and v = g (x) be functions that are continuous. Then, according to the work,
d (uv)) = udv + vdu or udv = d (uv) - vdu.
For the expression d (uv), the antiderivative will obviously be uv, so the following formula holds:
∫ udv = uv - ∫ vdu (8.4.)
This formula expresses the rule integration by parts... It brings the integration of the expression udv = uv "dx to the integration of the expression vdu = vu" dx.
Let, for example, it is required to find ∫xcosx dx. We put u = x, dv = cosxdx, so du = dx, v = sinx. Then
∫xcosxdx = ∫x d (sin x) = x sin x - ∫sin x dx = x sin x + cosx + C.
The rule of integration by parts has a more limited scope than variable substitution. But there are whole classes of integrals, for example,
∫x k ln m xdx, ∫x k sinbxdx, ∫ x k cosbxdx, ∫x k e ax and others, which are calculated using integration by parts.
Definite integral
The concept of a definite integral is introduced as follows. Let the function f (x) be defined on the segment. We split the segment [a, b] into n parts by points a = x 0< x 1 <...< x n = b. Из каждого интервала (x i-1 ,
x i) возьмем произвольную точку ξ i и составим сумму f(ξ i)
Δx i где
Δ x i = x i - x i-1. A sum of the form f (ξ i) Δ x i is called integral sum, and its limit as λ = maxΔx i → 0, if it exists and is finite, is called definite integral function f (x) of a before b and is indicated by:
F (ξ i) Δx i (8.5).
The function f (x) in this case is called integrable on the segment, numbers a and b are called the lower and upper limit of the integral.
The following properties are valid for a definite integral:
4), (k = const, k∈R);
5)
6)
7) f (ξ) (b-a) (ξ∈).
The last property is called mean value theorem.
Let f (x) be continuous on. Then on this segment there is an indefinite integral
∫f (x) dx = F (x) + C
and takes place Newton-Leibniz formula, connecting a definite integral with an indefinite one:
F (b) - F (a). (8.6)
Geometric interpretation: the definite integral is the area of a curved trapezoid bounded from above by the curve y = f (x), straight lines x = a and x = b and an axis segment Ox.
Improper integrals
Integrals with infinite limits and integrals of discontinuous (unbounded) functions are called improper. Improper integrals of the first kind - these are integrals over an infinite interval, defined as follows:
(8.7)
If this limit exists and is finite, then it is called convergent improper integral of f (x) on the interval [a, + ∞), and the function f (x) is called integrable on an infinite interval[a, + ∞). Otherwise, the integral is said to be does not exist or diverges.
Improper integrals on the intervals (-∞, b] and (-∞, + ∞) are defined similarly:
Let us define the concept of an integral of an unbounded function. If f (x) is continuous for all values x segment, except for the point c, at which f (x) has an infinite discontinuity, then improper integral of the second kind of f (x) ranging from a to b called the amount:
if these limits exist and are finite. Designation:
Examples of calculating integrals
Example 3.30. Calculate ∫dx / (x + 2).
Solution. We denote t = x + 2, then dx = dt, ∫dx / (x + 2) = ∫dt / t = ln | t | + C = ln | x + 2 | + C.
Example 3.31... Find ∫ tgxdx.
Solution.∫ tgxdx = ∫sinx / cosxdx = - ∫dcosx / cosx. Let t = cosx, then ∫ tgxdx = -∫ dt / t = - ln | t | + C = -ln | cosx | + C.
Example3.32 ... Find ∫dx / sinxSolution.
Example3.33. Find .
Solution. = .
Example3.34 ... Find ∫arctgxdx.
Solution. We integrate by parts. We set u = arctgx, dv = dx. Then du = dx / (x 2 +1), v = x, whence ∫arctgxdx = xarctgx - ∫ xdx / (x 2 +1) = xarctgx + 1/2 ln (x 2 +1) + C; because
∫xdx / (x 2 +1) = 1/2 ∫d (x 2 +1) / (x 2 +1) = 1/2 ln (x 2 +1) + C.
Example3.35 ... Calculate ∫lnxdx.
Solution. Applying the formula for integration by parts, we get:
u = lnx, dv = dx, du = 1 / x dx, v = x. Then ∫lnxdx = xlnx - ∫x 1 / x dx =
= xlnx - ∫dx + C = xlnx - x + C.
Example3.36 ... Evaluate ∫e x sinxdx.
Solution. Denote u = e x, dv = sinxdx, then du = e x dx, v = ∫sinxdx = - cosx → ∫ e x sinxdx = - e x cosx + ∫ e x cosxdx. The integral ∫e x cosxdx is also integrable by parts: u = e x, dv = cosxdx, du = e x dx, v = sinx. We have:
∫ e x cosxdx = e x sinx - ∫ e x sinxdx. We got the relation ∫e x sinxdx = - e x cosx + e x sinx - ∫ e x sinxdx, whence 2∫e x sinx dx = - e x cosx + e x sinx + С.
Example 3.37. Calculate J = ∫cos (lnx) dx / x.
Solution. Since dx / x = dlnx, then J = ∫cos (lnx) d (lnx). Replacing lnx by t, we arrive at the tabular integral J = ∫ costdt = sint + C = sin (lnx) + C.
Example 3.38 ... Calculate J =.
Solution. Considering that = d (lnx), we substitute lnx = t. Then J = 
.
Example 3.39
... Calculate the integral J = 
.
Solution. We have: 
... Therefore =
=
=. entered like this sqrt (tan (x / 2)).
And if you click on Show steps in the upper right corner in the result window, you will get a detailed solution.
Complex integrals
This article completes the topic of indefinite integrals, and includes integrals that I find quite difficult. The lesson was created at the repeated requests of visitors who expressed their wishes that more difficult examples were also analyzed on the site.
It is assumed that the reader of this text is well prepared and knows how to apply the basic techniques of integration. Dummies and people who are not very confident about integrals should refer to the very first lesson - Indefinite integral. Examples of solutions, where you can master the topic practically from scratch. More experienced students can familiarize themselves with the techniques and methods of integration that have not yet been encountered in my articles.
What integrals will be considered?
First, we will consider integrals with roots, for the solution of which we successively use variable replacement and integration by parts... That is, in one example, two techniques are combined at once. And even more.
Then we will get acquainted with an interesting and original the method of reducing the integral to itself... Not so few integrals are solved in this way.
The third number of the program will go to integrals of complex fractions, which flew past the box office in previous articles.
Fourth, additional integrals of trigonometric functions will be analyzed. In particular, there are methods that avoid the time-consuming universal trigonometric substitution.
(2) In the integrand, we divide the numerator by the denominator term by term.
(3) We use the linearity property of the indefinite integral. In the last integral immediately we bring the function under the differential sign.
(4) Take the remaining integrals. Note that parentheses can be used in the logarithm, not modulus, since.
(5) We carry out the reverse substitution, expressing from the direct substitution "te":
Masochistic students can differentiate the answer and get the original integrand as I just did. No, no, I did the check in the correct sense =)
As you can see, in the course of the solution, even more than two solution methods had to be used, thus, to deal with such integrals, confident integration skills and not the smallest experience are needed.
In practice, of course, the square root is more common, here are three examples for an independent solution:
Example 2
Find the indefinite integral
Example 3
Find the indefinite integral
Example 4
Find the indefinite integral
These examples are of the same type, so the complete solution at the end of the article will be only for Example 2, in Examples 3-4 - one answer. Which substitution to use at the beginning of the solutions, I think, is obvious. Why did I pick up examples of the same type? They often meet in their role. More often, perhaps, just something like 
.
But not always, when the root of a linear function is found under the arctangent, sine, cosine, exponent, and other functions, several methods have to be applied at once. In a number of cases, it is possible to "get off easily", that is, immediately after the replacement, a simple integral is obtained, which can be taken in an elementary way. The easiest of the tasks proposed above is Example 4, in which, after replacing, a relatively simple integral is obtained.
By reducing the integral to itself
An ingenious and beautiful method. Let's take a look at the classics of the genre immediately:
Example 5
Find the indefinite integral
There is a square binomial under the root, and when trying to integrate this example, the kettle can suffer for hours. Such an integral is taken piece by piece and reduced to itself. In principle, not difficult. If you know how.
Let us denote the integral under consideration by a Latin letter and start the solution: 
We integrate piece by piece: 
(1) Prepare an integrand function for term division.
(2) We divide the integrand by term. Perhaps not everyone understands, I will write in more detail:
(3) We use the linearity property of the indefinite integral.
(4) Take the last integral ("long" logarithm).
Now we look at the very beginning of the solution: 
And at the end: 
What happened? As a result of our manipulations, the integral reduced to itself!
Let's equate the beginning and the end: 
Move to the left with a sign change: 
And we carry the deuce to the right side. As a result: 
The constant, strictly speaking, should have been added earlier, but added it at the end. I strongly recommend that you read what is strict here:
Note:
More strictly, the final stage of the solution looks like this:
Thus:
The constant can be redesignated as. Why can you re-designate? Because it still accepts any values, and in this sense there is no difference between constants and.
As a result:
A similar constant redesignation trick is widely used in differential equations... And there I will be strict. And here such liberty is allowed by me only in order not to confuse you with unnecessary things and to focus on the very method of integration.
Example 6
Find the indefinite integral
Another typical integral for an independent solution. Complete solution and answer at the end of the tutorial. The difference with the answer from the previous example will be!
If there is a square trinomial under the square root, then the solution in any case is reduced to two analyzed examples.
For example, consider the integral 
... All you need to do is in advance select a full square:
.
Further, a linear replacement is carried out, which is dispensed with "without any consequences":
, resulting in an integral. Something familiar, right?
Or such an example, with a square binomial:
Select a complete square:
And, after a linear replacement, we get an integral, which is also solved according to the already considered algorithm.
Consider two more typical examples of how to reduce an integral to itself:
- integral of the exponent multiplied by the sine;
Is the integral of the exponent multiplied by the cosine.
In the listed integrals by parts, we will have to integrate two times already:
Example 7
Find the indefinite integral
The integrand is the exponent multiplied by the sine.
We integrate by parts twice and reduce the integral to itself: 
As a result of double integration by parts, the integral reduced to itself. Let's equate the beginning and the end of the solution:
Move to the left with a sign change and express our integral: 
Ready. Along the way, it is advisable to comb the right side, i.e. put the exponent outside the brackets, and in the brackets arrange the sine and cosine in a "nice" order.
Now let's go back to the beginning of the example, or rather to integration by parts: 
For we have designated the exhibitor. The question arises, exactly the exponent should always be denoted by? Not necessary. In fact, in the considered integral fundamentally does not matter What to denote for, it was possible to go the other way: 
Why is this possible? Because the exponent turns into itself (both during differentiation and integration), sine and cosine mutually transform into each other (again, both during differentiation and integration).
That is, you can also designate a trigonometric function. But, in the considered example, this is less rational, since fractions will appear. If you wish, you can try to solve this example in the second way, the answers must be the same.
Example 8
Find the indefinite integral
This is an example for a do-it-yourself solution. Before deciding, think about what is more profitable in this case to designate for, exponent or trigonometric function? Complete solution and answer at the end of the tutorial.
And of course, keep in mind that most of the answers in this lesson are easy enough to differentiate!
The examples were considered not the most difficult. In practice, integrals are more common, where the constant is both in the exponent and in the argument of the trigonometric function, for example:. Many people will have to get lost in such an integral, and I myself often get confused. The fact is that in the solution there is a high probability of the appearance of fractions, and it is very easy to lose something by inattention. In addition, there is a high probability of error in the signs, note that the exponent has a minus sign, and this introduces additional difficulty.
At the final stage, it often turns out something like the following:
Even at the end of the solution, you should be extremely careful and competently deal with fractions: 
Integration of compound fractions
We are slowly getting closer to the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complicated, just for one reason or another the examples were a little "off topic" in other articles.
Continuing the theme of roots
Example 9
Find the indefinite integral
In the denominator under the root is the square trinomial plus outside the root "appendage" in the form of "x". An integral of this kind is solved using a standard substitution.
We decide: 
The replacement is simple: 
We look at life after replacement: 
(1) After substitution, we bring the terms under the root to a common denominator.
(2) We take out from under the root.
(3) Reduce the numerator and denominator by. At the same time, under the root, I rearranged the terms in a convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions verbally.
(4) The resulting integral, as you remember from the lesson Integration of some fractions, solved full square selection method... Select a complete square.
(5) By integration we obtain an ordinary "long" logarithm.
(6) We carry out the reverse replacement. If initially, then back:.
(7) The final action is aimed at the hairstyle of the result: under the root, we again bring the terms to a common denominator and take them out from under the root.
Example 10
Find the indefinite integral 
This is an example for a do-it-yourself solution. Here, a constant has been added to the lonely X, and the replacement is almost the same: 
The only thing that needs to be done additionally is to express the "x" from the replacement: 
Complete solution and answer at the end of the tutorial.
Sometimes in such an integral there may be a square binomial under the root, this does not change the solution, it will be even simpler. Feel the difference:
Example 11
Find the indefinite integral
Example 12
Find the indefinite integral
Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution method of which was considered in the lesson Integrals of Irrational Functions.
Integral of an indecomposable polynomial of degree 2 in degree
(polynomial in the denominator)
More rare, but, nevertheless, encountered in practical examples, the form of the integral.
Example 13
Find the indefinite integral
But back to the example with the lucky number 13 (honestly, I didn't guess right). This integral is also from the category of those with which you can pretty much torment yourself if you don't know how to solve it.
The solution starts with an artificial transformation: 
I think everyone already understands how to divide the numerator by the denominator term by term.
The resulting integral is taken piece by piece: 
For an integral of the form (is a natural number), we have derived recurrent Degree reduction formula:
, where 
- integral of a degree lower.
Let us verify the validity of this formula for the solved integral.
In this case:,, we use the formula: 
As you can see, the answers are the same.
Example 14
Find the indefinite integral
This is an example for a do-it-yourself solution. The sample solution uses the above formula twice in succession.
If under the degree there is indecomposable square trinomial, then the solution is reduced to a binomial by selecting a complete square, for example:
What if there is an additional polynomial in the numerator? In this case, the method of undefined coefficients is used, and the integrand is expanded into the sum of fractions. But in my practice of such an example never met, so I skipped this case in the article Integrals of a fractional rational function, I will skip it now. If such an integral still occurs, see the textbook - everything is simple there. I do not consider it appropriate to include material (even simple ones), the probability of meeting with which tends to zero.
Integration of complex trigonometric functions
The adjective “difficult” for most examples is again largely conditional. Let's start with tangents and cotangents in high degrees. From the point of view of the methods used for solving the tangent and the cotangent, they are almost the same, so I will talk more about the tangent, implying that the demonstrated method for solving the integral is also valid for the cotangent too.
In the above lesson, we looked at universal trigonometric substitution for solving a certain kind of integrals of trigonometric functions. The disadvantage of the universal trigonometric substitution is that when using it, cumbersome integrals with difficult calculations often arise. And in some cases, universal trigonometric substitution can be avoided!
Consider another canonical example, the integral of unity divided by sine:
Example 17
Find the indefinite integral
Here you can use generic trigonometric substitution and get the answer, but there is a more rational way. I will provide a complete solution with comments for each step:
(1) We use the double angle sine trigonometric formula.
(2) We carry out an artificial transformation: In the denominator, divide and multiply by.
(3) According to the well-known formula in the denominator, we transform the fraction into a tangent.
(4) We bring the function under the sign of the differential.
(5) Take the integral.
A couple of simple examples for an independent solution:
Example 18
Find the indefinite integral
Hint: The very first step is to use the cast formula 
and carefully carry out the actions similar to the previous example.
Example 19
Find the indefinite integral
Well, this is a very simple example.
Complete solutions and answers at the end of the lesson.
I think now no one will have problems with integrals: 
etc.
What is the idea behind the method? The idea is to organize only the tangents and the derivative of the tangent in the integrand using transformations, trigonometric formulas. That is, we are talking about replacing: 
... In Examples 17-19, we actually applied this replacement, but the integrals were so simple that the matter was treated with an equivalent action - bringing the function under the differential sign.
Similar reasoning, as I have already mentioned, can be carried out for the cotangent.
There is also a formal prerequisite for applying the above replacement:
The sum of the powers of cosine and sine is a negative integer EVEN number, for example:
for an integral - a negative integer EVEN number.
! Note : if the integrand contains ONLY a sine or ONLY a cosine, then the integral is also taken for a negative odd degree (the simplest cases are in Examples No. 17, 18).
Consider a couple of more meaningful tasks for this rule:
Example 20
Find the indefinite integral
The sum of the powers of the sine and cosine: 2 - 6 = –4 is a negative integer EVEN number, which means that the integral can be reduced to tangents and its derivative: 
(1) Transform the denominator.
(2) According to the well-known formula, we obtain.
(3) Transform the denominator.
(4) We use the formula 
.
(5) We bring the function under the sign of the differential.
(6) We carry out a replacement. More experienced students may not carry out the replacement, but it is still better to replace the tangent with one letter - there is less risk of confusion.
Example 21
Find the indefinite integral
This is an example for a do-it-yourself solution.
Hold on, champion rounds begin =)
Often in the integrand there is a "hodgepodge":
Example 22
Find the indefinite integral 
This integral initially contains a tangent, which immediately prompts an already familiar thought:
Artificial transformation at the very beginning and the rest of the steps I will leave without comment, since everything has already been discussed above.
A couple of creative examples for self-solution:
Example 23
Find the indefinite integral 
Example 24
Find the indefinite integral 
Yes, in them, of course, you can lower the degrees of the sine, cosine, use the universal trigonometric substitution, but the solution will be much more efficient and shorter if you draw it through the tangents. Complete solution and answers at the end of the lesson
In the fifth century BC, the ancient Greek philosopher Zeno of Elea formulated his famous aporias, the most famous of which is the aporia "Achilles and the tortoise". This is how it sounds:Let's say Achilles runs ten times faster than a turtle and is a thousand steps behind it. During the time it takes Achilles to run this distance, the turtle will crawl a hundred steps in the same direction. When Achilles has run a hundred steps, the turtle will crawl ten more steps, and so on. The process will continue indefinitely, Achilles will never catch up with the turtle.
This reasoning came as a logical shock to all subsequent generations. Aristotle, Diogenes, Kant, Hegel, Hilbert ... All of them, in one way or another, considered Zeno's aporias. The shock was so strong that " ... discussions continue at the present time, the scientific community has not yet managed to come to a common opinion about the essence of paradoxes ... mathematical analysis, set theory, new physical and philosophical approaches were involved in the study of the issue; none of them has become a generally accepted solution to the question ..."[Wikipedia, Zeno's Aporia"]. Everyone understands that they are being fooled, but no one understands what the deception is.
From the point of view of mathematics, Zeno in his aporia clearly demonstrated the transition from magnitude to. This transition implies application instead of constants. As far as I understand, the mathematical apparatus for applying variable units of measurement either has not yet been developed, or it has not been applied to Zeno's aporia. Applying our usual logic leads us into a trap. We, by inertia of thinking, apply constant units of measurement of time to the reciprocal. From a physical point of view, it looks like time dilation until it stops completely at the moment when Achilles is level with the turtle. If time stops, Achilles can no longer overtake the turtle.
If we turn over the logic we are used to, everything falls into place. Achilles runs at a constant speed. Each subsequent segment of his path is ten times shorter than the previous one. Accordingly, the time spent on overcoming it is ten times less than the previous one. If we apply the concept of "infinity" in this situation, then it would be correct to say "Achilles will infinitely quickly catch up with the turtle."
How can you avoid this logical trap? Stay in constant time units and do not go backwards. In Zeno's language, it looks like this:
During the time during which Achilles will run a thousand steps, the turtle will crawl a hundred steps in the same direction. Over the next interval of time, equal to the first, Achilles will run another thousand steps, and the turtle will crawl a hundred steps. Now Achilles is eight hundred steps ahead of the turtle.
This approach adequately describes reality without any logical paradoxes. But this is not a complete solution to the problem. Einstein's statement about the insuperability of the speed of light is very similar to the Zeno aporia "Achilles and the Turtle". We still have to study, rethink and solve this problem. And the solution must be sought not in infinitely large numbers, but in units of measurement.
Another interesting aporia Zeno tells about a flying arrow:
A flying arrow is motionless, since at every moment of time it is at rest, and since it is at rest at every moment of time, it is always at rest.
In this aporia, the logical paradox is overcome very simply - it is enough to clarify that at each moment of time a flying arrow rests at different points in space, which, in fact, is motion. Another point should be noted here. From a single photograph of a car on the road, it is impossible to determine either the fact of its movement or the distance to it. To determine the fact of the movement of the car, two photographs are needed, taken from the same point at different points in time, but the distance cannot be determined from them. To determine the distance to the car, you need two photographs taken from different points in space at the same time, but it is impossible to determine the fact of movement from them (of course, additional data are still needed for calculations, trigonometry will help you). What I want to draw special attention to is that two points in time and two points in space are different things that should not be confused, because they provide different opportunities for research.
Wednesday, 4 July 2018
The distinction between set and multiset is very well described in Wikipedia. We look.
As you can see, "there cannot be two identical elements in a set", but if there are identical elements in a set, such a set is called a "multiset". Such logic of absurdity will never be understood by rational beings. This is the level of talking parrots and trained monkeys, who lack intelligence from the word "completely". Mathematicians act as ordinary trainers, preaching their absurd ideas to us.
Once the engineers who built the bridge were in a boat under the bridge during the tests of the bridge. If the bridge collapsed, the incompetent engineer died under the rubble of his creation. If the bridge could withstand the load, a talented engineer would build other bridges.
No matter how mathematicians hide behind the phrase "chur, I'm in the house", or rather "mathematics studies abstract concepts," there is one umbilical cord that inextricably connects them with reality. This umbilical cord is money. Let us apply mathematical set theory to the mathematicians themselves.
We studied mathematics very well and now we are sitting at the cash desk, giving out salaries. Here comes a mathematician for his money. We count the entire amount for him and lay out on our table into different piles, in which we put bills of the same denomination. Then we take one bill from each pile and hand the mathematician his “mathematical set of salary”. Let us explain the mathematics that he will receive the rest of the bills only when he proves that a set without identical elements is not equal to a set with identical elements. This is where the fun begins.
First of all, the logic of the deputies will work: "You can apply this to others, you can not apply to me!" Further, we will begin to assure us that there are different denomination numbers on bills of the same denomination, which means that they cannot be considered the same elements. Okay, let's count the salary in coins - there are no numbers on the coins. Here the mathematician will start to frantically remember physics: different coins have different amounts of dirt, the crystal structure and arrangement of atoms in each coin is unique ...
And now I have the most interesting question: where is the line beyond which the elements of a multiset turn into elements of a set and vice versa? Such a line does not exist - everything is decided by shamans, science did not lie anywhere near here.
Look here. We select football stadiums with the same pitch. The area of the fields is the same, which means we have got a multiset. But if we consider the names of the same stadiums, we get a lot, because the names are different. As you can see, the same set of elements is both a set and a multiset at the same time. How is it correct? And here the mathematician-shaman-shuller takes a trump ace out of his sleeve and begins to tell us either about the set or about the multiset. In any case, he will convince us that he is right.
To understand how modern shamans operate with set theory, tying it to reality, it is enough to answer one question: how do the elements of one set differ from the elements of another set? I'll show you, without any "thinkable as not a single whole" or "not thinkable as a whole."
Sunday, 18 March 2018
The sum of the digits of the number is a dance of shamans with a tambourine, which has nothing to do with mathematics. Yes, in mathematics lessons we are taught to find the sum of the digits of a number and use it, but that is why they are shamans in order to teach their descendants their skills and wisdom, otherwise shamans will simply die out.
Need proof? Open Wikipedia and try to find the Sum of Digits of a Number page. It doesn't exist. There is no formula in mathematics by which you can find the sum of the digits of any number. After all, numbers are graphic symbols with the help of which we write numbers and in the language of mathematics the task sounds like this: "Find the sum of graphic symbols representing any number". Mathematicians cannot solve this problem, but shamans - it is elementary.
Let's see what and how we do in order to find the sum of the digits of a given number. And so, let us have the number 12345. What should be done in order to find the sum of the digits of this number? Let's go through all the steps in order.
1. We write down the number on a piece of paper. What have we done? We have converted the number to the graphic symbol of the number. This is not a mathematical operation.
2. We cut one resulting picture into several pictures containing separate numbers. Cutting a picture is not a mathematical operation.
3. Convert individual graphic symbols to numbers. This is not a mathematical operation.
4. Add up the resulting numbers. Now that's mathematics.
The sum of the digits of 12345 is 15. These are the "cutting and sewing courses" from shamans used by mathematicians. But that is not all.
From the point of view of mathematics, it does not matter in which number system we write the number. So, in different number systems, the sum of the digits of the same number will be different. In mathematics, the number system is indicated as a subscript to the right of the number. With a large number 12345, I do not want to fool my head, consider the number 26 from the article about. Let's write this number in binary, octal, decimal and hexadecimal number systems. We will not look at every step under a microscope, we have already done that. Let's see the result.
As you can see, in different number systems, the sum of the digits of the same number is different. This result has nothing to do with mathematics. It’s the same as if you would get completely different results when you determined the area of a rectangle in meters and centimeters.
Zero in all number systems looks the same and has no sum of digits. This is another argument for the fact that. A question for mathematicians: how is something that is not a number designated in mathematics? What, for mathematicians, nothing but numbers exists? For shamans, I can allow this, but for scientists - no. Reality is not all about numbers.
The result obtained should be considered as proof that number systems are units of measurement for numbers. After all, we cannot compare numbers with different units of measurement. If the same actions with different units of measurement of the same quantity lead to different results after their comparison, then this has nothing to do with mathematics.
What is real mathematics? This is when the result of a mathematical action does not depend on the magnitude of the number, the unit of measurement used and on who performs this action.
Ouch! Isn't this a women's toilet?
- Young woman! This is a laboratory for the study of the indiscriminate holiness of souls during the ascension to heaven! Halo on top and up arrow. What other toilet?
Female ... The nimbus above and the down arrow is male.
If a piece of design art like this flashes before your eyes several times a day,
Then it is not surprising that in your car you suddenly find a strange icon:
Personally, I make an effort on myself so that in a pooping person (one picture), I can see minus four degrees (a composition of several pictures: minus sign, number four, degrees designation). And I don’t think this girl is a fool who doesn’t know physics. She just has a stereotype of perception of graphic images. And mathematicians constantly teach us this. Here's an example.
1A is not "minus four degrees" or "one a". This is "pooping man" or the number "twenty six" in hexadecimal notation. Those people who constantly work in this number system automatically perceive the number and the letter as one graphic symbol.
