Approximate values ​​of real numbers. Approximate values ​​of real numbers Approximate calculation of function values

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Let it be required to find with accuracy up to (with disadvantage). Let's arrange the calculations like this:

We first find the approximate root, accurate to 1, only from the integer 2. We get 1 (and the remainder is 1). We write the number 1 at the root and put a comma after it. Now we find the number of tenths. To do this, we add to the remainder 1 the numbers 3 and 5, located to the right of the decimal point, and continue the extraction as if we were extracting the root of the integer 235. We write the resulting number 5 in the root in the place of tenths. We don't need the remaining digits of the radical number (104). That the resulting number 1.5 will actually be an approximate root to within , can be seen from the following; if we were to find the largest integer root of 235 with an accuracy of 1, we would get 15, which means

Dividing each of these numbers by 100, we get:

(By adding the number 0.00104, the double sign ≤ should obviously change to the sign<, а знак >remains (since 0.00104< 0,01).)

Suppose we want to find an approximate one with a disadvantage, up to an accuracy. Let's find the whole number, then the tenths figure, then the hundredths figure. The root of an integer is 15 integers. To get the tenths digit, we must, as we have seen, add two more digits to the remainder 23, to the right of the decimal point:

In our example, these numbers are not present at all; put zeros in their place. By adding them to the remainder and continuing as if we were finding the root of the integer 24800, we will find the tenths figure 7. It remains to find the hundredths figure. To do this, we add two more zeros to the remainder 151 and continue extraction, as if we were finding the root of the integer 2480000. We get 15.74. That this number is really an approximate root of 248 with an accuracy of up to a disadvantage can be seen from the following. If we were to find the largest integer square root of the integer 2480000, we would get 1574, which means

Dividing each of these numbers by 10000 (1002), we get:

15,74 2 ≤ 248; 15,75 2 > 248.

This means that 15.74 is that decimal fraction that we called an approximate root with a disadvantage with an accuracy of up to 248.

Rule. To extract from a given integer or from a given decimal fraction an approximate root with a deficiency with an accuracy of the root has 0 integers).

Then they find the number of tenths. To do this, add two digits of the conquered number to the right of the decimal point to the remainder (if they are not there, add two zeros to the remainder), and continue extraction as is done when extracting the root of an integer. The resulting number is written at the root in the place of tenths.

Then find the hundredths number. To do this, two numbers to the right of those that were just removed are added to the remainder, etc.

Thus, when extracting the root of an integer with a decimal fraction the number must be divided into edges with two digits each, starting from the decimal point, both to the left (in the integer part of the number) and to the right (in the fractional part).

Examples.

In the last example, we converted a fraction to a decimal by calculating eight decimal places to create the four faces needed to find the four decimal places of the root.

Let it be necessary to calculate a definite integral $\int\limits_(a)^(b)f(x)dx$ with some predetermined accuracy $\varepsilon$. If directly finding the antiderivative integrand function $f(x)$ is too cumbersome, or the integral $\int f(x)dx$ is not taken at all, then in these cases you can use functional series. In particular, Maclaurin series are used, with the help of which a power series expansion of the integrand of the function $f(x)$ is obtained. That is why in our work we will need a document with Maclaurin series.

The power series that we will use converge uniformly, so they can be integrated term by term over any segment lying inside the interval of convergence. The scheme for solving similar problems involving the calculation of integrals using series is simple:

  1. Expand the integrand into a functional series (usually a Maclaurin series).
  2. Carry out term-by-term integration of the terms of the functional series written in the first paragraph.
  3. Calculate the sum of the number series obtained in the second step with the specified accuracy $\varepsilon$.

Problems involving the calculation of integrals using series are popular among compilers of standard calculations in higher mathematics. Therefore, in this topic we will analyze five examples, in each of which we need to calculate a definite integral with an accuracy of $\varepsilon$.

Example No. 1

Calculate $\int\limits_(0)^(\frac(1)(2))e^(-x^2)dx$ accurate to $\varepsilon=10^(-3)$.

Let us immediately note that the integral $\int e^(-x^2)dx$ is not taken, i.e. the antiderivative of the integrand is not expressed through a finite combination of elementary functions. In other words, it will not be possible to find the antiderivative of the function $e^(-x^2)$ using standard methods (substitution, integration by parts, etc.).

There are two design options for such tasks, so we will consider them separately. Conventionally, they can be called “expanded” and “shortened” versions.

Expanded design option

Maclaurin series:

$$e^x=1+x+\frac(x^2)(2)+\frac(x^3)(6)+\ldots$$

$$e^(-x^2)=1-x^2+\frac(\left(-x^2\right)^2)(2)+\frac(\left(-x^2\right) ^3)(6)+\ldots=1-x^2+\frac(x^4)(2)-\frac(x^6)(6)+\ldots$$

We integrate the resulting expansion on the interval $\left$:

$$\int\limits_(0)^(\frac(1)(2))e^(-x^2)dx=\int\limits_(0)^(\frac(1)(2))\left (1-x^2+\frac(x^4)(2)-\frac(x^6)(6)+\ldots\right)dx=\\ =\left.\left(x-\frac( x^3)(3)+\frac(x^5)(10)-\frac(x^7)(42)+\ldots\right)\right|_(0)^(1/2)= \ frac(1)(2)-\frac(1)(3\cdot(2^3))+\frac(1)(10\cdot(2^5))-\frac(1)(42\cdot( 2^7))+\ldots$$

We have obtained a convergent sign of an alternating series. This means that if, to calculate the approximate value of a given integral, we take $k$ terms of the resulting series, then the error will not exceed the modulus of the $(k+1)$th term of the series.

According to the condition, the accuracy is $\varepsilon=10^(-3)$. Since $\frac(1)(42\cdot(2^7))=\frac(1)(5376)<10^{-3}$, то для достижения требуемой точности достаточно ограничиться первыми тремя членами знакочередующегося ряда:

$$\int\limits_(0)^(\frac(1)(2))e^(-x^2)dx\approx\frac(1)(2)-\frac(1)(3\cdot( 2^3))+\frac(1)(10\cdot(2^5))=\frac(443)(960).$$

The error of the resulting equality does not exceed $\frac(1)(5376)$.

However, summing ordinary fractions is a tedious task, so most often calculations are carried out in decimal fractions:

$$\int\limits_(0)^(\frac(1)(2))e^(-x^2)dx\approx\frac(1)(2)-\frac(1)(3\cdot( 2^3))+\frac(1)(10\cdot(2^5))\approx(0(,)5)-0(,)0417+0(,)0031\approx(0(,)461 ).$$

Of course, in this case rounding error must be taken into account. The first term (i.e. $0(,)5$) was calculated exactly, so there is no rounding error there. The second and third terms were taken rounded to the fourth decimal place, therefore the rounding error for each of them will not exceed $0.0001$. The resulting rounding error will not exceed $0+0(,)0001+0(,)0001=0(,)0002$.

Consequently, the total error of the equality $\int\limits_(0)^(\frac(1)(2))e^(-x^2)dx\approx(0(,)461)$ will not exceed $0(,)0002 +\frac(1)(5376)<10^{-3}$, т.е. значение интеграла вычислено с требуемой точностью.

Shortened version of the design

Let us write the expansion of the function $e^x$ in the Maclaurin series:

$$e^x=\sum\limits_(n=0)^(\infty)\frac(x^n)(n$$ !}

This expansion is valid for all $x\in(R)$. Let's substitute $-x^2$ instead of $x$:

$$e^(-x^2)=\sum\limits_(n=0)^(\infty)\frac(\left(-x^2\right)^n)(n=\sum\limits_{n=0}^{\infty}\frac{(-1)^n\cdot{x}^{2n}}{n!}$$ !}

We integrate the resulting series on the interval $\left$:

$$\int\limits_(0)^(\frac(1)(2))e^(-x^2)dx=\int\limits_(0)^(\frac(1)(2))\sum \limits_(n=0)^(\infty)\frac((-1)^n\cdot(x)^(2n))(ndx= \sum\limits_{n=0}^{\infty}\frac{(-1)^n}{n!}\int\limits_{0}^{\frac{1}{2}}x^{2n}dx=\\ =\sum\limits_{n=0}^{\infty}\frac{(-1)^n}{n!}\left.\frac{x^{2n+1}}{2n+1}\right|_{0}^{1/2}= \sum\limits_{n=0}^{\infty}\frac{(-1)^n\cdot\left(\frac{1}{2}\right)^{2n+1}}{n!\cdot(2n+1)}= \sum\limits_{n=0}^{\infty}\frac{(-1)^n}{n!\cdot(2n+1)\cdot{2^{2n+1}}}$$ !}

$$\sum\limits_(n=0)^(\infty)\frac((-1)^n)(n!\cdot(2n+1)\cdot(2^(2n+1)))=\ frac(1)(2)-\frac(1)(24)+\frac(1)(320)-\frac(1)(5376)+\ldots$$

All considerations that were made regarding errors in the expanded version of the design remain valid, i.e. $\int\limits_(0)^(\frac(1)(2))e^(-x^2)dx\approx\frac(1)(2)-\frac(1)(3\cdot(2 ^3))+\frac(1)(10\cdot(2^5))\approx(0(,)461)$.

Why is the shortened version of the recording better than the expanded one?

Firstly, we do not need to guess how many terms of the series to take in the original expansion in order to calculate the definite integral with a given accuracy. For example, we wrote down at the very beginning of the solution:

$$e^(-x^2)=1-x^2+\frac(x^4)(2)-\frac(x^6)(6)+\ldots$$

However, why did we decide that we need to take exactly four terms of the series? What if you need to take two members of a series, or five, or a hundred? If only the sixth term of the series turned out to be less than $\varepsilon$, what then? And then we would have to go back to the very beginning of the solution, add a couple more terms of the series and integrate them. And if this is not enough, then do this procedure again.

The abbreviated form of recording does not suffer from such a drawback. We get a number series written in general form, so we can take as many of its members as needed.

Based on the above reasons, I prefer the shortened recording method. In the future, all decisions in this topic will be presented in an abbreviated form.

Answer: $\int\limits_(0)^(\frac(1)(2))e^(-x^2)dx\approx(0(,)461)$.

Example No. 2

Calculate the definite integral $\int\limits_(0)^(0(,)2)\frac(1-\cos\frac(5x)(3))(x)dx$ accurate to $\varepsilon=10^( -3)$, expanding the integrand into a Maclaurin series and integrating term by term.

Let's start by expanding the integrand function $\frac(1-\cos\frac(5x)(3))(x)$ into a Maclaurin series. Let us write the expansion of the function $\cos(x)$ in the Maclaurin series:

$$\cos(x)=\sum\limits_(n=0)^(\infty)\frac((-1)^n\cdot(x)^(2n))((2n)$$ !}

This expansion is valid for all $x\in(R)$. Let's substitute the fraction $\frac(5x)(3)$ instead of $x$:

$$\cos(\frac(5x)(3))=\sum\limits_(n=0)^(\infty)\frac((-1)^n\cdot(\left(\frac(5x)( 3)\right))^(2n))((2n)= \sum\limits_{n=0}^{\infty}\frac{(-1)^n\cdot{5^{2n}}\cdot{x}^{2n}}{3^{2n}\cdot{(2n)!}}.$$ !}

Now let's expand $1-\cos\frac(5x)(3)$:

$$ 1-\cos\frac(5x)(3)=1-\sum\limits_(n=0)^(\infty)\frac((-1)^n\cdot(5^(2n))\ cdot(x)^(2n))(3^(2n)\cdot((2n)} $$ !}

Taking from the sum $\sum\limits_(n=0)^(\infty)\frac((-1)^n\cdot(5^(2n))\cdot(x)^(2n))(3^( 2n)\cdot((2n)}$ первый член, получим: $\sum\limits_{n=0}^{\infty}\frac{(-1)^n\cdot{5^{2n}}\cdot{x}^{2n}}{3^{2n}\cdot{(2n)!}}=1+\sum\limits_{n=1}^{\infty}\frac{(-1)^n\cdot{5^{2n}}\cdot{x}^{2n}}{3^{2n}\cdot{(2n)!}}$. Следовательно:!}

$$ 1-\sum\limits_(n=0)^(\infty)\frac((-1)^n\cdot(5^(2n))\cdot(x)^(2n))(3^( 2n)\cdot((2n)}=1-\left(1+\sum\limits_{n=1}^{\infty}\frac{(-1)^n\cdot{5^{2n}}\cdot{x}^{2n}}{3^{2n}\cdot{(2n)!}}\right)=\\ =-\sum\limits_{n=1}^{\infty}\frac{(-1)^n\cdot{5^{2n}}\cdot{x}^{2n}}{3^{2n}\cdot{(2n)!}} =\sum\limits_{n=1}^{\infty}\frac{-(-1)^n\cdot{5^{2n}}\cdot{x}^{2n}}{3^{2n}\cdot{(2n)!}}=\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}\cdot{5^{2n}}\cdot{x}^{2n}}{3^{2n}\cdot{(2n)!}}. $$ !}

The last thing left is to divide by $x$:

$$ \frac(1-\cos\frac(5x)(3))(x)=\frac(1)(x)\cdot\sum\limits_(n=1)^(\infty)\frac(( -1)^(n+1)\cdot(5^(2n))\cdot(x)^(2n))(3^(2n)\cdot((2n)}= \sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}\cdot{5^{2n}}\cdot{x}^{2n-1}}{3^{2n}\cdot{(2n)!}}. $$ !}

Let's integrate this expansion on the interval $\left$:

$$ \int\limits_(0)^(0(,)2)\frac(1-\cos\frac(5x)(3))(x)dx=\int\limits_(0)^(\frac( 1)(5))\sum\limits_(n=1)^(\infty)\frac((-1)^(n+1)\cdot(5^(2n))\cdot(x)^(2n -1))(3^(2n)\cdot((2n)}dx= \sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}\cdot{5^{2n}}}{3^{2n}\cdot{(2n)!}}\int\limits_{0}^{\frac{1}{5}}{x}^{2n-1}dx=\\ =\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}\cdot{5^{2n}}}{3^{2n}\cdot{(2n)!}}\cdot\left.\frac{x^{2n}}{2n}\right|_{0}^{1/5}= \sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{{2n}\cdot 3^{2n}\cdot{(2n)!}} $$ !}

We received a sign for an alternating row. Let's write down the first few terms of this series (until the written term becomes less than $\varepsilon$):

$$\sum\limits_(n=1)^(\infty)\frac((-1)^(n+1))((2n)\cdot 3^(2n)\cdot((2n)}=\frac{1}{36}-\frac{1}{7776}+\ldots$$ !}

Since $\frac(1)(7776)<\varepsilon$, то для вычисления интеграла с точностью $\varepsilon$ достаточно первого члена полученного числового ряда:

$$\int\limits_(0)^(0(,)2)\frac(1-\cos\frac(5x)(3))(x)dx\approx\frac(1)(36)\approx( 0(,)028).$$

Answer: $\int\limits_(0)^(0(,)2)\frac(1-\cos\frac(5x)(3))(x)dx\approx(0(,)028)$.

We will continue the topic of calculating integrals using Maclaurin series in

Power series are widely used in approximate calculations. With their help, you can calculate the values ​​of roots, trigonometric functions, logarithms of numbers, and definite integrals with a given accuracy. Series are also used when integrating differential equations.

Approximate calculation of function values

Consider the expansion of a function in a power series:

In order to calculate the approximate value of a function at a given point X, belonging to the region of convergence of the indicated series, the first ones are left in its expansion n members ( n– a finite number), and the remaining terms are discarded:

To estimate the error of the obtained approximate value, it is necessary to estimate the discarded remainder r n(x). To do this, use the following techniques:

- if the resulting series is alternating, then the following property is used: for an alternating series that satisfies the Leibniz conditions, the remainder of the series in absolute value does not exceed the first discarded term.

If a given series is of constant sign, then the series composed of discarded terms is compared with an infinitely decreasing geometric progression.

In the general case, to estimate the remainder of the Taylor series, you can use the Lagrange formula: (or x ).

Example 1 . Using the series expansion sin x, calculate sin20 o with an accuracy of 0.0001.

Solution. To be able to use formula (2), it is necessary to express the value of the argument in radian measure. We get . Substituting this value into the formula, we get

The resulting series is alternating in sign and satisfies Leibniz's conditions. Because , then this and all subsequent terms of the series can be discarded, limiting ourselves to the first two terms. Thus,

Example 2 . Calculate to the nearest 0.01.

Solution. Let's use the expansion where (see example 5 in the previous topic):

Let's check whether we can discard the remainder after the first three terms of the expansion; to do this, we will evaluate it using the sum of an infinitely decreasing geometric progression:

.

So we can discard this remainder and get

.

Example 3 . Calculate to the nearest 0.0001.

Solution. Let's use the binomial series. Since 5 3 is the cube of an integer closest to 130, it is advisable to represent the number 130 as 130 = 5 3 +5.

since already the fourth term of the resulting alternating series satisfying the Leibniz criterion is less than the required accuracy:

, so it and the terms following it can be discarded.

Approximate calculation of definite integrals

Many practically necessary definite or improper integrals cannot be calculated using the Newton-Leibniz formula, because its application is associated with finding the antiderivative, which often does not have an expression in elementary functions. It also happens that finding an antiderivative is possible, but it is unnecessarily labor-intensive. However, if the integrand function is expanded into a power series, and the limits of integration belong to the interval of convergence of this series, then an approximate calculation of the integral with a predetermined accuracy is possible.

Example 4 : Calculate the integral to the nearest 0.00001.

Solution. The corresponding indefinite integral cannot be expressed in elementary functions, i.e. represents a “non-permanent integral”. The Newton-Leibniz formula cannot be applied here. Let's calculate the integral approximately.

Dividing term by term the series for sin x on x, we get:

Integrating this series term by term (this is possible, since the limits of integration belong to the interval of convergence of this series), we obtain:

Since the resulting series satisfies Leibniz’s conditions and it is enough to take the sum of the first two terms to obtain the desired value with a given accuracy.

Thus, we find

.

Example 5 . Calculate the integral to the nearest 0.001.

Let's check whether we can discard the remainder after the second term of the resulting series.

Hence, .

Approximate solution of the Cauchy problem for ordinary

Differential equation

In frequent cases when an ODE cannot be solved in general form, the Cauchy problem for it can be solved approximately, in the form of the first few terms of the Taylor series expansion of the solution (in the neighborhood of a given point)

Example Find the first 3 terms of the series expansion of the solution to the Cauchy problem

Solution: We will look for a solution to the problem in the form

Coefficient at(1)=2 is the initial condition of the Cauchy problem.

We will find the coefficient from the equation by substituting the initial conditions into it:

Let's differentiate both sides of this equation to find:

Thus,

Decide : Calculate approximately with the specified accuracy:

A 1) up to 0.0001 2) up to 0.0001 3) up to 0.01 4) ln6 up to 0.01

5) up to 0.001 6) up to 0.001 7) up to 0.01

8) up to 0.001 9) up to 0.001 10) up to 0.001

11) up to 0.001 12) up to 0.01 13) up to 0.001

14) up to 0.001 15) up to 0.001 16) up to 0.001

B Find the first few terms of the series expansion of the solution to the Cauchy problem:

17) y¢-4y+xy 2 -e 2 x =0; y(0)=2 (4 terms) 18) y¢+ycosx-y 2 sinx=0; y(p)=1 (4 terms)

19) y¢¢=e y cozy¢; y(1)=1; y¢(1)=p/6 (5 terms)

20) y¢¢=xy 2 -1/y¢; y(0)=0, y¢(0)=1 (5 terms)

Fourier series

Near Fourier functions f(x) on the interval (-p;p)

, Where

Near Fourier functions f(x) on the interval (-l;l) is called a trigonometric series of the form:

, Where

Fourier series piecewise continuous, piecewise monotonic and bounded on the interval (- l;l) of the function converges on the entire number line.

Sum of Fourier series S(x):

Is a periodic function with period 2 l

On the interval (- l;l) coincides with the function f(x), except for breakpoints

At discontinuity points (of the first kind, since the function is limited) the functions f(x) and at the ends of the interval takes average values:

They say that a function expands into a Fourier series on the interval (- l;l): .

If f(x) is an even function, then only even functions participate in its expansion, that is b n=0.

If f(x) is an odd function, then only odd functions participate in its expansion, that is and n=0

Near Fourier functions f(x) on the interval (0;l) by cosines of multiple arcs the row is called:

, Where .

Near Fourier functions f(x) on the interval (0;l) by the sines of multiple arcs the row is called:

, Where .

The sum of the Fourier series over the cosines of multiple arcs is an even periodic function with period 2 l, coinciding with f(x) on the interval (0; l) at points of continuity.

The sum of the Fourier series over the sines of multiple arcs is an odd periodic function with period 2 l, coinciding with f(x) on the interval (0; l) at points of continuity.

The Fourier series for a given function on a given interval has the property of uniqueness, that is, if the expansion is obtained in some other way than using formulas, for example, by selecting coefficients, then these coefficients coincide with those calculated from the formulas.

Examples.

1. Expand the function f(x)=1:

a) in a complete Fourier series on the interval(-p;p) ;

b) in a series along the sines of multiple arcs on the interval(0;p); plot the resulting Fourier series

Solution:

a) The Fourier series expansion on the interval (-p;p) has the form:

,

and all coefficients b n=0, because this function is even; Thus,

Obviously, the equality will be satisfied if we accept

A 0 =2, A 1 =A 2 =A 3 =…=0

Due to the uniqueness property, these are the required coefficients. Thus, the required decomposition: or just 1=1.

In this case, when a series identically coincides with its function, the graph of the Fourier series coincides with the graph of the function on the entire number line.

b) The expansion on the interval (0;p) in terms of the sines of multiple arcs has the form:

It is obviously impossible to select the coefficients so that equality holds identically. Let's use the formula to calculate the coefficients:

Thus, for even n (n=2k) we have b n=0, for odd ( n=2k-1) -

Finally, .

Let's plot the resulting Fourier series using its properties (see above).

First of all, we build a graph of this function on a given interval. Next, taking advantage of the oddness of the sum of the series, we continue the graph symmetrically to the origin:

On September 9, 2007, driver Logan Gomez won the Chicagoland 100 of the IRL Indy Pro Series. He beat the second place winner by 0.0005 seconds, setting a record for tight finish in world motorsport. What equipment allows you to measure time with such accuracy?

On the wave of the lighthouse In modern racing, timing is completely automatic. Each car is equipped with a radio beacon that emits radio waves at a unique frequency. Antennas located in strictly defined places on the track pick up its signal and determine by frequency which car has passed by. The antennas are arranged two side by side: by measuring the time it takes to travel the distance from one antenna to another, the computer determines the speed of the vehicle. Up to 20 antennas can be located on the route. Special antennas are used to control speed in the pit lane. Information from radio receivers goes to the timing center, where more than 20 engineers continuously monitor the operation of computers. Just in case, the timing system is duplicated by a pair of infrared photocells installed at the finish line

Tim Skorenko

It is in the Indycar series that the timing requirements are the most stringent. No other championship can boast of measuring time with an accuracy of ten thousandths of a second. The overwhelming number of series is limited to 0.001 s, and this is most often enough with a reserve, but there are also incidents: for example, at the qualification of the 1997 European Grand Prix in the Formula 1 class, as many as three pilots managed to show a time that coincided to a thousandth of a second, - 1.21.072. Pole position eventually went to Jacques Villeneuve, who completed his fastest lap before the others.

In Formula 1, timing accuracy has varied markedly over time. In the first championship in 1950, 0.1 s was enough to fully record the finish of the pilots. There was not a single race included in the championship standings where the gap between the drivers was less than a second. Accuracy to 0.1 dates back to the very first Grand Prix in the history of motor racing - the French Grand Prix of 1906, where the time of the winner, Ferenc Schisz in a Renault, was 12 hours 14 minutes and 7.4 seconds (unmatched by the short and easy today's races, right?). In most races held before the First World War, the accuracy did not exceed 1 second.


In modern racing, timing is completely automatic. Each car is equipped with a radio beacon that emits radio waves at a unique frequency. Antennas located in strictly defined places on the track pick up its signal and determine by frequency which car has passed by. The antennas are arranged two side by side: by measuring the time it takes to travel the distance from one antenna to another, the computer determines the speed of the vehicle. Up to 20 antennas can be located on the route. Special antennas are used to control speed in the pit lane. Information from radio receivers goes to the timing center, where more than 20 engineers continuously monitor the operation of computers. Just in case, the timing system is duplicated by a pair of infrared photocells installed at the finish line.

In America, timekeepers were much more progressive. Post-war AAA racing (later CART) most often required measurement accuracy of up to 0.01. This was primarily due to the configuration of the tracks and the abundance of ovals, where the gaps between drivers are extremely small. The incredible timing accuracy of modern IRL is due to the same factor: of the seventeen rounds of the 2010 championship, eight are held on ovals.

Incidents and failures

Auto racing timing is inextricably linked with the world's leading watch and electronics manufacturers: TAG Heuer, Tissot, Omega, Longines... Almost all of them are represented in various sports as official timekeepers. Errors and inaccuracies in time measurement are practically excluded today. From 1992 to this day, the aforementioned European Grand Prix '97 has become the only chronometric curiosity of Formula 1, and in the IRL even such incidents are completely impossible.


Today, Indycar and NASCAR timing systems are considered among the best in the world. Each track is equipped in such a way that European organizers can only envy. The count goes by 0.0001 seconds (for Indycar), and live viewers at any time can get information about the speed of each car on the track, its lap time and any of the lap sectors, gaps in the pelaton with an accuracy of the sector, etc. d. - in general, maximum information. In a race where half of the season takes place on ovals, precise timing plays a huge role. The winner is often determined by a photo finish.

Oddly enough, the concept of “official timekeeper” has only recently appeared. It is today that Tissot “leads” the world motorcycle racing championship, and no other company has the right to interfere. Just 30 years ago, each individual race had its own timekeepers, “armed” with the equipment that the organizers could purchase.

Before World War II, in almost all racing series and classes, timing was carried out manually: specially trained people with stopwatches stood at the track. They recorded the lap time of the next car and recorded the data. However, there were also “breakthroughs”. In 1911, at the first Indianapolis 500, engineer Charlie Warner designed and implemented the first semi-automatic timing system in history. A thin wire was loosely stretched along the start-finish line and slightly raised above the brick surface. Each machine pressed the wire to the ground, increasing its tension. A stamping hammer was attached to the wire, which, when pulled, placed an ink mark on a slowly crawling graduated tape. The measurement accuracy reached 0.01 s! The timekeeper manually set the car numbers opposite each point. The system did not take root for a funny reason: in the middle of the race, driver Herb Little's car broke a wire. While they pulled on the new one (running in front of speeding cars), at least 20 laps passed, during which the timing was kept approximately. The race victory was awarded to Ray Harrown in Marmon, but another famous driver, Ralph Mulford, was convinced until his death that he had won the first-ever Indy 500.


The successful use of semi-automatic systems flourished in the 1930s. The Indy 500 back then used Stewart-Warner or huge Loughborough-Hayes chronographs.

In the early years of the NASCAR series, timing was absolutely terrible. In some races, a person sat at the finish line with a paper and pencil and recorded: so-and-so is first, so-and-so is second. True, this only applied to gravel and mud tracks. Things were better at the racetracks. In particular, it was at the Elkhart Lake race in 1951 that the Streeter-Amet chronograph was used. The device sequentially printed (in tenths of a second) on a paper tape the time of each passing car; the person’s job was to write car numbers opposite each number.

The fully automatic timing system was first used in the USAC Championship race at Ontario Speedway in 1970. Each car was equipped with a transmitter that emitted waves at its own, unique frequency. An antenna was installed at the start-finish line that picked up the oscillation frequency of each transmitter; the rest of the work was done by the computer.


Professional timekeeper David McKinney, who worked at various races in Australia and New Zealand in the 1960s, gave us interesting information: “If the most skilled timekeeper with the best chronometer can exactly 'catch' a tenth of a second, he is simply lucky.” all manual measurements ever taken in racing can safely be considered approximate.

"Formula 1"

In Europe, automatic systems appeared much later than in America. In international series like Formula 1, confusion and vacillation reigned. Until the late 1970s, timing at different Grand Prix events was handled by completely different people, using different equipment and methods. At free races, the role of timekeepers was most often performed by the wives of the racers. For example, Norma Hill, the wife of two-time world champion Graham Hill, went with her husband to every Grand Prix and personally timed his lap times, double-checking the work of the marshals.


In the mid-1970s, tired of constant confusion and mistakes, the Ferrari team began bringing its own high-precision equipment purchased in America to the Grand Prix. One of the mechanics of Ferrari's archrival Lotus asked his boss Colin Chapman: "Why don't we do the same?" “Do you really think this will make our cars go faster?” - Chapman answered. This answer very accurately characterizes the European attitude to timekeeping accuracy in those years. However, by the end of the 1970s, almost all major teams entered into contracts with watch manufacturers and carried their own timing systems with them. After one of the races, Autosport magazine wrote: “Teams publish timings of such precision in official reports that the official figures of the Grand Prix organizers look like they were made using a Mickey Mouse clock!”

Remarkable incidents regularly occurred due to timing errors. For example, during the rainy Canadian Grand Prix of 1973, a safety car was brought onto the track for the first time. The timekeepers were confused, confused with the lap times and incorrectly added up the times before and after the pace car. As a result, the victory was successively celebrated by Emerson Fittipaldi from Lotus, Jackie Oliver from Shadow and Peter Revson from McLaren. The victory went to the latter - after several hours of bickering.


An equally interesting story happened at the 1975 Swedish Grand Prix. March rider Vittorio Brambilla was far from the fastest in the pelaton, but it was he who won pole position in that race. This was because March designer Robin Heard had quietly passed directly in front of the photocell of the recording instrument half a second before Brambilla crossed the finish line. By some miracle, no one saw this, and the device recorded the time of Heard on foot, and not the racer at all.

A triumph of technology

Today's racing is a celebration of high technology. For example, the NASCAR series was almost the last to switch to modern timing methods, adhering to traditions as much as possible. But today NASCAR's timing systems are considered some of the best in the world. Tissot, the official timekeeper of the overseas series for the past four years, has equipped each track in a way that European organizers can only envy. In a race where 34 of the season's 36 rounds are held on ovals, precise timing plays a huge role.


No less serious systems are used in the world motorcycle racing championship (Tissot is also its timekeeper). Unlike NASCAR, there is no need for complex monitoring systems to determine who is ahead: the motorcyclists are not in such a dense field. But since the MotoGP tracks are of a traditional European configuration, and not ovals, there are also plenty of difficulties. Setting time cutoffs at certain places on the route requires careful thought (ovals are simply geometrically divided into 4-8 parts).

Today's computer technology virtually eliminates the possibility of timing errors in auto or motorcycle racing. The organizers of the Grand Prix have long found completely different problems on their minds - safety, ecology, etc. And the time fixers work for themselves and work. You could say like a clock.

Let it be required to find Y 2.35104 with an accuracy of (with a disadvantage). Let's arrange the calculations like this:

We first find the approximate root with an accuracy of 1 only from the integer 2. We get 1 (and the remainder is 1). We write the number 1 at the root and put a comma after it. Now we find the number of tenths. To do this, we add to the remainder 1 the numbers 3 and 5, located to the right of the decimal point, and continue the extraction as if we were extracting the root of the integer 235. We write the resulting number 5 in the root in the place of tenths. We don't need the remaining digits of the radical number (104). That the resulting number 1.5 will really be an approximate root, with an accuracy up to the successor; if

we found the largest integer root of 235 with an accuracy of 1, then we would get 15, which means

Dividing each of these numbers by 100, we get;

in finally

Suppose you want to find an approximate one with a disadvantage, up to an accuracy. Let's find the whole number, then the tenths figure, then the hundredths figure. The root of an integer is 15 integers. To get the tenths digit, we must, as we have seen, add two more digits to the remainder 23, to the right of the decimal point:

In our example, these numbers are not present at all; put zeros in their place. By adding them to the remainder and continuing as if we were finding the root of the integer 24,800, we will find the tenths figure 7. It remains to find the hundredths figure. To do this, we add two more zeros to the remainder 151 and continue extraction, as if we were finding the root of the integer 2 480000. We get 15.74. That this number is really an approximate root of 248 with an accuracy of up to a disadvantage can be seen from the following. If we were to find the largest integer square root of the integer 2,480,000, we would get 1574, which means

Dividing each of these numbers by 10,000 (100^2), we get:

This means that 15.74 is that decimal fraction that we called an approximate root with a disadvantage accurate to within 248.

Rule. To extract from a given integer or from a given decimal fraction an approximate root with a deficiency accurate to up to to, etc., first find an approximate root with a deficiency accurate to 1 by extracting the root from the integer (if it is not there, write in the root 0 integers).

Then they find the number of tenths. To do this, add two digits of the radical number to the right of the decimal point to the remainder (if they are not there, add two zeros to the remainder), and continue extraction as is done when extracting a root from an integer. The resulting number is written at the root in the place of tenths.

Then find the hundredths number. To do this, two numbers to the right of those that were just removed are added to the remainder, etc.

Thus, when extracting the root of an integer with a decimal fraction, the number must be divided into edges of two digits each, starting from the decimal point, both to the left (in the integer part of the number) and to the right (in the fractional part).

1. Extract exactly to the roots:

2. Extract with precision

In the last example, we converted the fraction y to a decimal by calculating eight decimal places to form the four faces needed to find the four decimal places of the root.

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