Some tasks of school mathematics. Part II

Part I. Fractions
Part II. Modules

This article discusses the method of estimating the range of accepted values ​​and the relationship of this method with the tasks containing the module.

When solving some problems, it is necessary to consider the range within which the desired value may be.
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Consider the method of estimation when solving inequalities.

Suppose that the price for a single item can vary from 5 to 10 RUB. To give a top estimate means to determine the maximum value that the desired value can take. For two units of goods, the price for which does not exceed 10, the upper rating will be 10 + 10 = 20 .

Consider the task from the taskbook of the profile orientation of M.I. Bashmakova
37. Known estimates for variables $$$x$and $$$y: 0 <x <5, 2 <y <3.$

Give top estimates for the following expressions:
one. $$$2x + 3y$
2 $$$xy$


five. $$$\ frac {1} {y}$
6 $$$\ frac {x} {y}$

eight. $$$xy$
9. $$$3x-2y$


In general, the analysis of infinitely small quantities uses an evaluation criterion. The concept of a module as a neighborhood lies in the definition of the limit itself.

$$

$$\ left | x_ {n} -a \ right | <\ varepsilon$$

Consider an example from the Course of Differential and Integral Calculus 363 (6)

It is easy to establish the divergence of the series

$$

$$\ sum \ frac {1} {\ sqrt {n}} = 1+ \ frac {1} {\ sqrt {2}} + \ frac {1} {\ sqrt {3}} + ... + \ frac {1} {\ sqrt {n}} + ...$$

In fact, since its members decrease, the nth partial sum

$$

$$1+ \ frac {1} {\ sqrt {2}} + ... + \ frac {1} {\ sqrt {n}}> n \ cdot \ frac {1} {\ sqrt {n}} = \ sqrt {n}$$

and grows to infinity along with $$$n$.

In order to prove that $$$1+ \ frac {1} {\ sqrt {2}} + ... + \ frac {1} {\ sqrt {n}}$really more $$$\ sqrt {n}$, you need to evaluate the bottom of this expression. We get a system of inequalities

$$

$$\ left \ {\! \ begin {aligned} & \ frac {1} {\ sqrt {n-1}}> \ frac {1} {\ sqrt {n}} \\ & \ frac {1} {\ sqrt {n-2}}> \ frac {1} {\ sqrt {n}} \\ & \ frac {1} {\ sqrt {n-3}}> \ frac {1} {\ sqrt {n}} \\ & ... \ end {aligned} \ right.$$

Making the addition of all the inequalities of this system, we get

$$

$$1+ \ frac {1} {\ sqrt {2}} + \ frac {1} {\ sqrt {3}} + ... + \ frac {1} {\ sqrt {n}}> \ frac {1 } {\ sqrt {n}} + \ frac {1} {\ sqrt {n}} + \ frac {1} {\ sqrt {n}} + ... + \ frac {1} {\ sqrt {n} } = n \ cdot \ frac {1} {\ sqrt {n}}$$

This is proof that this series diverges.

With a harmonic series, such reception does not work, because $$$n$-th partial sum of the harmonic series

$$

$$1+ \ frac {1} {2} + \ frac {1} {3} + ... + \ frac {1} {n}> n \ cdot \ frac {1} {n} = 1$$

Let's return to the task

38. Calculate the amount ("Tasks for children from 5 to 15 years")

$$

$$\ frac {1} {1 \ cdot2} + \ frac {1} {2 \ cdot3} + \ frac {1 {3 \ cdot4} + ... + \ frac {1} {99 \ cdot100}$$

(with an error of not more than 1% of the answer)

Estimate from above sum of series $$$\ frac {n} {n + 1}$gives the number 1.

Drop the first addend $$$\ frac {1} {1 \ cdot2}$

(define series_sum_1 ( lambda (n) (if (= n 0) 0 (+ (/ 1.0 (* (+ n 1.0 )(+ n 2.0))) (series_sum_1(- n 1.0))) ) ) ) (writeln (series_sum_1 10)) (writeln (series_sum_1 100)) (writeln (series_sum_1 1000)) (writeln (series_sum_1 10000)) (writeln (series_sum_1 100000)) (writeln (series_sum_1 1000000)) 

Will get $$$1 - \ frac {1} {1 \ cdot2} = \ frac {1} {2}$
0.41666666666666663
0.49019607843137253
0.4990019960079833
0.4999000199960005
0.49999000019998724
0.4999990000019941

You can check it at ideone.com here.


Drop the first two terms. $$$\ frac {1} {1 \ cdot2} + \ frac {1} {2 \ cdot3}$

 (define series_sum_1 ( lambda (n) (if (= n 0) 0 (+ (/ 1.0 (* (+ n 2.0) (+ n 3.0))) (series_sum_1(- n 1.0))) ) ) ) (series_sum_1 1000000) 

We get 0.33333233333632745
Partial sums of a series are bounded above.

A positive row always has an amount; this sum will be finite (and, therefore, the row will converge) if the partial sums of the row are bounded above, and infinite (and the row divergent) otherwise.

Throw away $$$n$the initial components of the harmonic series.
Prove (using the lower bound) that

$$

$$\ frac {1} {n + 1} + \ frac {1} {n + 2} + ... + \ frac {1} {2n}> \ frac {1} {2}$$

If, discarding the first two members, the remaining members of the harmonic series are divided into groups by $$$2,4,8, ..., 2 ^ {k-1}, ...$members in each

$$

$$\ frac {1} {3} + \ frac {1} {4}; \ frac {1} {5} + \ frac {1} {6} + \ frac {1} {7} + \ frac {1} {8}; \ frac {1} {9} + ... \ frac {1} {16}; ...;$$

$$

$$\ frac {1} {2 ^ {k-1} +1} + ... + \ frac {1} {2 ^ {k}}; ...,$$

then each of these amounts separately will be more $$$\ frac {1} {2}$.
... We see that partial sums cannot be bounded above: the series has an infinite sum.

Calculate the partial amounts that are obtained by dropping $$$2 ^ {k}$terms.

 #lang racket (* 1.0 (+ 1/3 1/4)) (* 1.0 (+ 1/5 1/6 1/7 1/8)) (* 1.0 (+ 1/9 1/10 1/11 1/12 1/13 1/14 1/15 1/16)) 

We get:
0.5833333333333334
0.6345238095238095
0.6628718503718504
We write a program that calculates the sum of the harmonic series from $$$\ frac {n} {2}$before $$$n$where $$$n = 2 ^ {k}$at $$$k \ in \ mathbb {N}$

 #lang racket (define (Hn n ) (define half_arg (/ n 2.0)) (define (series_sum n) (if (= n half_arg ) 0 (+ (/ 1.0 n) (series_sum(- n 1)) ) ) ) (series_sum n) ) (Hn 4) (Hn 8) (Hn 16) (Hn 32) 

We get:
0.5833333333333333
0.6345238095238095
0.6628718503718504
0.6777662022075267

You can check in the online ide link
For range $$$\ left [1 + 2 ^ {70}; 2 ^ {71} \ right]$we get 0.693147 ...
Check out the Wolfram Cloud here .

This recursive algorithm causes a fast stack overflow.
This article has an example of factorial calculation using an iterative algorithm. We modify this iterative algorithm so that it calculates the partial sum Hn within certain bounds; let's call these boundaries a and b

 (define (Hn ab) (define (iteration product counter) (if (> counter b) product (iteration (+ product (/ 1.0 counter)) (+ counter 1)))) (iteration 0 a)) 

The bottom is the number $$$1 + 2 ^ {k}$, the upper bound is the number $$$2 \ cdot 2 ^ {k}$
We write a function that calculates the power of two

 (define (power_of_two k) (define (iteration product counter) (if (> counter k) product (iteration (* product 2) (+ counter 1)))) (iteration 1 1)) 

We will substitute (+ 1 (power_of_two k)) as the lower border, and use the function (* 2 (power_of_two k)) or its equivalent function (power_of_two (+ 1 k)) as the top border
Rewrite the function Hn

 (define (Hn k) (define a (+ 1 (power_of_two k)) ) (define b (* 2 (power_of_two k)) ) (define (iteration product counter) (if (> counter b) product (iteration (+ product (/ 1.0 counter)) (+ counter 1)))) (iteration 0 a )) 

Now you can calculate Hn for large values $$$k$.

We write in C language a program that measures the time needed to calculate Hn . We will use the clock () function from the standard library <time.h>
Article about measurement of processor time is on Habré here .

 #include <math.h> #include <stdio.h> #include <time.h> int main(int argc, char **argv) { double count; // 1+2^30 -> 2^31 for(unsigned long long int i=1073741825 ;i<=2147483648 ;i++) { count=count+1.0/i; } printf("Hn = %.12f ", count); double seconds = clock() / (double) CLOCKS_PER_SEC; printf("  %f  \n", seconds); return 0; } 

Let's return to the modules.
In integral calculus, the module is used in the formula

$$

$$\ int \ frac {1} {x} dx = \ int \ frac {dx} {x} = ln \ left | x \ right | + C$$

On Habré there was an article The most natural logarithm in which this integral is considered and based on its calculation of the number $$$e$.

The presence of the module in the formula $$$\ int \ frac {dx} {x} = ln \ left | x \ right | + C$further substantiated in the "Course of differential and integral calculus"

If a… $$$x <0$then it’s easy to make sure that $$$\ left [ln (-x) \ right] '= \ frac {1} {x}$

Physical integral application $$$\ int \ frac {dx} {x}$

This integral is used to calculate the potential difference between the plates of a cylindrical capacitor.


"Electricity and Magnetism":

The potential difference between the plates we find by integrating:

$$

$$\ varphi_ {1} - \ varphi_ {2} = \ int \ limits_ {R_ {1}} ^ {R_ {2}} E (r) dr = \ frac {q} {2 \ pi \ varepsilon_ {0} \ varepsilon l} \ int \ limits_ {R_ {1}} ^ {R_ {2}} \ frac {dr} {r} = \ frac {q} {2 \ pi \ varepsilon_ {0} \ varepsilon l} ln \ frac {R_ {2}} {R_ {1}}$$

( $$$R_ {1}$and $$$R_ {2}$- the radii of the inner and outer plates).

The module sign under the sign of the natural logarithm is not used here. $$$ln \ left | \ frac {R_ {2}} {R_ {1}} \ right |$, because $$$R_ {1}$and $$$R_ {2}$strictly positive and this form of writing is redundant.

"Modular" drawing

With the help of modules you can draw various shapes.

If the program geogebra write the formula $$$abs (x) + abs (y) = 1$will get



You can draw more complex shapes. For example, let's draw a “butterfly” in the WolframAlpha cloud.

$$

$$\ sum \ frac {\ left | x \ right | } {n- \ left | x \ right | } + \ frac {\ left | x + n \ right | } {n} + \ frac {\ left | x-n \ right | } {n}$$

Plot [Sum [abs (x) / (n-abs (x)) + abs (x + n) / (n) + abs (xn) / (n), {n, 1.20}], {x, -60,60}]
In this expression $$$n$lies in the range of $$$1$before $$$20$, $$$x$lies in the range of $$$-60$before $$$60$.
Link to the picture.

Books:

"The task book of the profile orientation" M.I. Bashmakov
The course of general physics: in 3 tons. T. 2. “Electricity and magnetism” I.V. Saveliev