Some interesting methods of solving functional equations | Статья в журнале «Молодой ученый»

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Автор:

Рубрика: Математика

Опубликовано в Молодой учёный №32 (270) август 2019 г.

Дата публикации: 10.08.2019

Статья просмотрена: 4 раза

Библиографическое описание:

Саматбоева М. Т. Some interesting methods of solving functional equations // Молодой ученый. — 2019. — №32. — С. 1-5. — URL https://moluch.ru/archive/270/61843/ (дата обращения: 17.09.2019).



The article presents different ways of solving functional equations that occur in olimpiads and contests.

В статье представлены разные способы решения функциональных равенств, которые встречаются на олимпиадах и на конкурсах.

  1. Find all functions such that

is nonzero and holds for all [1].

Solution: Let be the assertion .

From we get

: and

, so

By subtracting the second equation from the first we get

(*). Now: Let be the assertion

We have , so

Again by subtracting the second equation from the first we get: Now for ,

We know natural divisors of are Then must be equal one of these . So we consider eight cases:

1) Suppose .

In (*) from , is clear and such prime number is not exist.

2) Suppose . From (*), . is clear and in we have such prime number is not exist.

3) Suppose . From (*),

In , Contradiction. is satisfied. is not satisfied.

For , such prime number only

4) Suppose . In (*), For any prime ,

. such prime number is not exist.

5) . Such as can not be.

6)Suppose . In (*), But for any prime ,

. such prime number is not exist

7) . Such as can not be.

So for any , Now from

. From this we get

By subtracting the second equation from the first we get: And also we have

. By subtracting these we get

Now assume . Then and for prime numbers, From this But for enough large - prime number this is not true, because, such - prime number infinite. Contradiction. So we conclude, such is not exist. For ,

  1. Find all functions such that holds for all [2].

Solution: be the assertion .

: Now consider this set . Values of elements of are bounded, lies between and . Let the smallest. We have . . But because of the smallest, so for . Now assume . : Also we have In this for (

Then . this number is positive and costanta, but is unbounded. Contradiction. So we get . From this equality we conclude the biggest element’s value and the smallest element’s value of are equal. All element’s value of equal to .

. : and

toq

For , from this,

So, But unbounded, but is positive Contradiction. So, such is not exist. , This is indeed a solution.

  1. Find all functions such that for all

[3].

Solution: be the assertion .

: (here ) .

:

From this Now for , (**).

: .

: . : : From (**).

: For . ( is clear). for all . This is indeed a solution

  1. Find all functions for all

[3].

Solution: be the assertion .

: injective and : . In this equality : .

: Because of injective: or If , from Contradiction We have the following equalities:

:

: (in (**) :

). Because of injective:

. In this equality : . From (***): . we can write (***):

: (****) : .

.

  1. Find all functions for all [1].

.

Solution: be the assertion (*).

:

: In this equality . using this equality we can rewrite (*).

.

: : 2. In this equality :

Now using this and (**) we get

. So . (****). Assume : (: ). from (***)

: But from (****) or Contradiction. Such is not exist. ,

  1. Find all functions such that for the following equalities hold:

, .

Yechim: be the assertion .

:

: (*).

Let . -case: suppose . In (*) :

: : Contradiction.

-case: suppose . In (*) : So, If then from we get . So If , according to second condition, .

-case: . : . :

: Using this and (**), for , (***).Using this we can rewrite initial equality:

from (***) For

: So,

In (****) : , From (***), for ,

References:

  1. www.artofproblemsolving.com/community/c482986. IMO Shortlisted problems 2016.
  2. Mohammad Mahdi Taheri. “Functional equations in mathematical competitions: Problems and solutions” July 1, 2015.
  3. Ozgur Kircak. “Functional equations” April 8, 2011.


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