March 2, 2019 at 09:35

Tasks from school textbook II

    Part I
    Part II

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

    When solving some problems, it is necessary to consider the range within which the desired value can be.

    Consider the estimation method for solving inequalities.

    Suppose that the price per unit of goods can range from 5 to 10 RUB. To give an upper bound means to determine the maximum value that the desired quantity can take. For two units of goods, the price for which does not exceed 10, the upper estimate will be 10 + 10 = 20 .

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

    Give top marks for the following expressions:
    1.$ 2x + 3y $
    2. $ xy $
    Guide to solving problems 5 and 6
    To evaluate fractional expressions, it is necessary to use the following property of numerical inequalities:
    • If $ a <b $ and both numbers are positive, then $ \ frac {1} {a}> \ frac {1} {b} $


    5. $ \ frac {1} {y} $
    6. $ \ frac {x} {y} $
    Instructions for solving problems 8 and 9
    To evaluate negative values, it is necessary to use the following property of numerical inequalities:
    If$ a <b $ and both numbers are positive, then $ -a> -b $

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


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

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



    Consider the example from the "Course of differential and integral calculus" 363 (6)
    Easy to set row divergence

    $ \ 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 ad infinitum with $ n $.
    In order to prove that $ 1+ \ frac {1} {\ sqrt {2}} + ... + \ frac {1} {\ sqrt {n}} $ really more $ \ sqrt {n} $, you need to make a lower estimate of this expression. We obtain the 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. $


    After adding 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 range, this technique does not work, because$ n $partial harmonic series

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



    We return to task

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

    $ \ 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)

    An upper estimate of the sum of the series$ \ frac {n} {n + 1} $gives the number 1.

    Drop the first term$ \ 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))

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

    You can check in ideone.com here

    The same algorithm in Python
    def series_sum(n):
	if n==0:
		return 0
	else:
		return 1.0/((n+1.0)*(n+2.0))+series_sum(n-1.0)
print(series_sum(10))
print(series_sum(100))

    Link to ideone.com

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

    We calculate the sum of the harmonic series with increasing n
    #lang racket
(define series_sum_1
 ( lambda (n)
  (if (= n 0) 0 
    (+ (/ 1.0 n) (series_sum_1(- n 1.0)))
  ) ) )
(series_sum_1 10)
(series_sum_1 100)
(series_sum_1 1000)
(series_sum_1 10000)
(series_sum_1 100000)
(series_sum_1 1000000)

    We
    get : 2.9289682539682538
    5.187377517639621
    7.485470860550343
    9.787606036044348
    12.090146129863335
    14.392726722864989


    Throw away $ n $initial terms 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 terms, 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 individually will be larger $ \ frac {1} {2} $.
    ... We see that partial sums cannot be bounded above: the series has an infinite sum.

    We calculate the partial amounts that are obtained by discarding $ 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
    Let's 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 : 0.583333333333333333
    0.6345238095238095
    0.6628718503718504
    0.6777662022075267

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

    This recursive algorithm causes a fast stack overflow.
    In this article there is an example of the factorial using an iterative algorithm. We modify this iterative algorithm so that it calculates the partial sum Hn within certain boundaries; call these boundaries a and b
    (define (Hn a b)
  (define (iteration product counter)
    (if (> counter b)
        product 
        (iteration (+ product (/ 1.0 counter))
                   (+ counter 1))))
  (iteration 0 a))

    The lower bound 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 boundary, and use the function (* 2 (power_of_two k)) or its equivalent function (power_of_two (+ 1 k)) as the upper boundary. 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 a program that measures the time required to calculate Hn . We will use the clock () function from the standard library
    An article about measuring processor time is on Habré here .
    #include 
#include 
#include 
int main(int argc, char **argv)
{
double count; 
// k от 1+2^30 до 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("Программа работала %f сек \n", seconds);
 return 0;
}

    Обычно online ide ограничивают время выполнения запускаемых программ пятью секундами, поэтому эту программу можно проверить лишь в некоторых online ide, например, в onlinegdb.com или repl.it
    При k от 1+2^30 до 2^31 время работы составит ~5 сек.
    При k от 1+2^31 до 2^32 время работы составит ~10 сек.
    При k от 1+2^32 до 2^33 время работы составит ~20 сек.
    При k от 1+2^33 до 2^34 время работы составит ~40 сек.
    При k от 1+2^34 до 2^35 время работы составит больше одной минуты.
    …
    При k от 1+2^45 до 2^46 время работы составит больше 24 часов.

    Предположим, что для k от 1+2^30 до 2^31 время выполнения алгоритма составляет ~2 сек.
    Тогда для k=2^(30+n) время выполнения алгоритма составляет 2^n сек. (при $ n \ in \ mathbb {N} $)
    Данный алгоритм имеет экспотенциальную сложность.

    Вернёмся к модулям.
    В интегральном исчислении модуль используется в формуле

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



    На Хабре была статья Самый натуральный логарифм, в которой рассматривается этот интеграл и на основе его вычисление числа $ e $.

    Присутствие модуля в формуле $ \ int \ frac {dx} {x} = ln \ left | x \ right | + C $ обосновывается далее в «Курсе дифференциального и интегрального исчисления»
    Если… $ x <0 $, то дифференцированием легко убедиться в том, что $ \ left [ln (-x) \ right] '= \ frac {1} {x} $

    Физическое приложение интеграла $ \ int \ frac {dx} {x} $


    Этот интеграл используется для вычисления разности потенциалов обкладок цилиндрического конденсатора.


    «Электричество и магнетизм»:
    Разность потенциалов между обкладками находим путем интегрирования:

    $ \ 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} $ и $ R_ {2} $ — радиусы внутренней и внешней обкладок).
    The module sign is not used here under the sign of the natural logarithm $ ln \ left | \ frac {R_ {2}} {R_ {1}} \ right | $, because $ R_ {1} $ and $ R_ {2} $ strictly positive and this form of recording is redundant.

    "Modular" drawing


    Using modules, you can draw various shapes.

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



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

    $ \ sum \ frac {\ left | x \ right | } {n- \ left | x \ right | } + \ frac {\ left | x + n \ right | } {n} + \ frac {\ left | xn \ 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 from $ 1 $ before $ 20 $, $ x $ lies in the range from $ -60 $ before $ 60 $.
    Link to the picture.

    Books:


    “The task book of profile orientation” M.I. Bashmakov
    General physics course: in 3 volumes T. 2. “Electricity and magnetism” I.V. Savelyev
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