New way to introduce exhibitors

The article proposes a new, very unusual way to determine the exponent and on the basis of this definition its main properties are derived.

Every positive number

a

$a$ set in correspondence

E_a = \ left \ {x: x = \ left (1 + a_1 \ right) \ left (1 + a_2 \ right) \ ldots \ left (1 + a_k \ right) \ right.

$E_a = \ left \ {x: x = \ left (1 + a_1 \ right) \ left (1 + a_2 \ right) \ ldots \ left (1 + a_k \ right) \ right.$ where

a_{1}, a_{2}, l d o t s, a_{k} > 0

$a_1, a_2, \ ldots, a_k> 0$ and

l e f t . a_{1} + a_{2} + l d o t s + a_{k} = a r i g h t}

$\ left.a_1 + a_2 + \ ldots + a_k = a \ right \}$ .

Lemma 1 . Of $0 <a <b$ it follows that for each element $x \ in E_a$ there is an item $y \ in E_b$ such that $y> x$ .

Will write

A l e q c

$A \ leq c$ , if a

c

$c$ upper bound of the set

A

$A$ . Similarly, we will write

A g e q c

$A \ geq c$ , if a

c

$c$ - lower bound of the set

A

$A$ .

Lemma 2. If $a_1, a_2, \ ldots, a_k> 0$ then $\ left (1 + a_1 \ right) \ left (1 + a_2 \ right) \ ldots \ left (1 + a_k \ right) \ geq 1 + a_1 + a_2 + \ text {...} + a_k$ .

Evidence

We proceed by induction.

For

k = 1

$k = 1$ The statement is obvious:

1 + a_{1} g e q 1 + a_{1}

$1 + a_1 \ geq 1 + a_1$ .

Let be

l e f t (1 + a_{1} r i g h t) l d o t s l e f t (1 + a_{i} r i g h t) g e q 1 + a_{1} + l d o t s + a_{i}

$\ left (1 + a_1 \ right) \ ldots \ left (1 + a_i \ right) \ geq 1 + a_1 + \ ldots + a_i$ for

1 < i < k

$1 <i <k$ .

Then

l e f t (1 + a_{1} r i g h t) l d o t s l e f t (1 + a_{i} r i g h t) l e f t (1 + a_{i + 1} r i g h t) g e q 1 + a_{1} + l d o t s + a_{i} + l e f t (1 + a_{1} + l d o t s + a_{i} r i g h t) a_{i + 1} g e q

$\ left (1 + a_1 \ right) \ ldots \ left (1 + a_i \ right) \ left (1 + a_ {i + 1} \ right) \ geq 1 + a_1 + \ ldots + a_i + \ left (1 + a_1 + \ ldots + a_i \ right) a_ {i + 1} \ geq$

g e q 1 + a_{1} + l d o t s + a_{i} + a_{i + 1}

$\ geq 1 + a_1 + \ ldots + a_i + a_ {i + 1}$ .

Lemma 2 is proved.

In the following, we show that each set

E_{a}

$E_a$ limited. Lemma 2 implies that

s u p E_{a} g e q a

$\ sup E_a \ geq a$ (one)

Lemma 3. If $0 <a \ leq \ frac {1} {2}$ and $a_1, \ ldots, a_k> 0$ , $a_1 + a_2 + \ ldots + a_k = a$ then $\ left (1 + a_1 \ right) \ ldots \ left (1 + a_i \ right) \ leq 1+ (1 + 2a) a_1 + (1 + 2a) a_2 + \ ldots + (1 + 2a) a_i$ , $i = 1,2, \ ldots, k$ .

Evidence

Indeed, by induction

1 + a_{1} l e q 1 + (1 + 2 a) a_{1}

$1 + a_1 \ leq 1+ (1 + 2a) a_1$ .

Let it be proved that

l e f t (1 + a_{1} r i g h t) l d o t s l e f t (1 + a_{i} r i g h t) l e q 1 + (1 + 2 a) a_{1} + l d o t s + (1 + 2 a) a_{i}

$\ left (1 + a_1 \ right) \ ldots \ left (1 + a_i \ right) \ leq 1+ (1 + 2a) a_1 + \ ldots + (1 + 2a) a_i$ .

Then

l e f t (1 + a_{1} r i g h t) l d o t s l e f t (1 + a_{i} r i g h t) l e f t (1 + a_{i + 1} r i g h t) l e q 1 + (1 + 2 a) a_{1} + l d o t s + (1 + 2 a) a_{i} +

$\ left (1 + a_1 \ right) \ ldots \ left (1 + a_i \ right) \ left (1 + a_ {i + 1} \ right) \ leq 1+ (1 + 2a) a_1 + \ ldots + (1 + 2a) a_i +$

+ l e f t (1 + (1 + 2 a) a_{1} + l d o t s + (1 + 2 a) a_{i} r i g h t) a_{i + 1} l e q

$+ \ left (1+ (1 + 2a) a_1 + \ ldots + (1 + 2a) a_i \ right) a_ {i + 1} \ leq$

l e q 1 + (1 + 2 a) a_{1} + l d o t s + (1 + 2 a) a_{i} + l e f t (1 + 2 a_{1} + l d o t s + 2 a_{i} r i g h t) a_{i + 1} l e q

$\ leq 1+ (1 + 2a) a_1 + \ ldots + (1 + 2a) a_i + \ left (1 + 2a_1 + \ ldots + 2a_i \ right) a_ {i + 1} \ leq$

l e q 1 + (1 + 2 a) a_{1} + l d o t s + (1 + 2 a) a_{i} + (1 + 2 a) a_{i + 1}

$\ leq 1+ (1 + 2a) a_1 + \ ldots + (1 + 2a) a_i + (1 + 2a) a_ {i + 1}$ .

Lemma 3 is proved.

Lemma 3 implies

Lemma 4. If $0 <a \ leq \ frac {1} {2}$ and $a_1, \ ldots, a_k> 0$ , $a_1 + a_2 + \ ldots + a_k = a$ then $\ left (1 + a_1 \ right) \ ldots \ left (1 + a_k \ right) \ leq 1 + a + 2a ^ 2$ .

Lemmas 3 and 4 imply an important inequality: if

0 < a l e q f r a c 12

$0 <a \ leq \ frac {1} {2}$ then

1 + a l e q E_{a} l e q 1 + a + 2 a^{2}

$1 + a \ leq E_a \ leq 1 + a + 2a ^ 2$ (2)

In particular, if

a l e q f r a c 12

$a \ leq \ frac {1} {2}$ then

E_{a} l e q 2

$E_a \ leq 2$ . Note that inequality

1 + a l e q E_{a}

$1 + a \ leq E_a$ true to all

a > 0

$a> 0$ .

Lemma 5. For any natural $n$ fair inequality $E_n \ leq 2 ^ {2n}$ .

Evidence

Let be

a_{1}, l d o t s, a_{k} > 0

$a_1, \ ldots, a_k> 0$ ,

a_{1} + l d o t s + a_{k} = n

$a_1 + \ ldots + a_k = n$ .

Evaluate the work

l e f t (1 + a_{1} r i g h t) l d o t s l e f t (1 + a_{k} r i g h t)

$\ left (1 + a_1 \ right) \ ldots \ left (1 + a_k \ right)$ . Lemma 2 implies that

l e f t (1 + f r a c a_{i} 2 n r i g h t)^{2 n} g e q 1 + a_{i}

$\ left (1+ \ frac {a_i} {2n} \ right) {} ^ {2n} \ geq 1 + a_i$ for

i = 1, l d o t s, k

$i = 1, \ ldots, k$ .

therefore

l e f t (1 + a_{1} r i g h t) l d o t s l e f t (1 + a_{k} r i g h t) l e q l e f t (1 + f r a c a_{1} 2 n r i g h t)^{2 n} l d o t s l e f t (1 + f r a c a_{k} 2 n r i g h t)^{2 n} = l e f t (l e f t (1 + f r a c a_{1} 2 n r i g h t) l d o t s l e f t (1 + f r a c a_{k} 2 n r i g h t) r i g h t)^{2 n}

$\ left (1 + a_1 \ right) \ ldots \ left (1 + a_k \ right) \ leq \ left (1+ \ frac {a_1} {2n} \ right) {} ^ {2n} \ ldots \ left ( 1+ \ frac {a_k} {2n} \ right) {} ^ {2n} = \ left (\ left (1+ \ frac {a_1} {2n} \ right) \ ldots \ left (1+ \ frac {a_k } {2n} \ right) \ right) {} ^ {2n}$ .

Because

f r a c a_{1} 2 n + l d o t s + f r a c a_{k} 2 n = f r a c 12

$\ frac {a_1} {2n} + \ ldots + \ frac {a_k} {2n} = \ frac {1} {2}$ , then applying Lemma 4, we obtain

l e f t (1 + f r a c a_{1} 2 n r i g h t) l d o t s l e f t (1 + f r a c a_{k} 2 n r i g h t) l e q 1 + f r a c 12 + 2 c d o t f r a c 14 = 2

$\ left (1+ \ frac {a_1} {2n} \ right) \ ldots \ left (1+ \ frac {a_k} {2n} \ right) \ leq 1+ \ frac {1} {2} +2 \ cdot \ frac {1} {4} = 2$ i.e.

l e f t (l e f t (1 + f r a c a_{1} 2 n r i g h t) l d o t s l e f t (1 + f r a c a_{k} 2 n r i g h t) r i g h t)^{2 n} l e q 2^{2 n}

$\ left (\ left (1+ \ frac {a_1} {2n} \ right) \ ldots \ left (1+ \ frac {a_k} {2n} \ right) \ right) {} ^ {2n} \ leq 2 ^ {2n}$ .

Thus, Lemma 5 is proved.

Lemma 6. Let $A \ subset B$ two non-empty bounded subsets of the set of real numbers $R$ . If for any $b \ in B$ there is an item $a \ in A$ such that $a \ geq b$ then $\ sup A = \ sup B$ .

Evidence

It's clear that

s u p A l e q s u p B

$\ sup A \ leq \ sup B$ . Assuming that

s u p A < s u p B

$\ sup A <\ sup B$ there is

v a r e p s i l o n > 0

$\ varepsilon> 0$ such that

t e x t s u p A < t e x t s u p B - v a r e p s i l o n

$\ text {sup} A <\ text {sup} B- \ varepsilon$ . So for any

a i n A

$a \ in A$ true inequality

a < s u p B - v a r e p s i l o n

$a <\ sup B- \ varepsilon$ . But in

B

$B$ there is an item

b > s u p B - v a r e p s i l o n

$b> \ sup B- \ varepsilon$ . So each

a i n A

$a \ in A$ less of this

b

$b$ , which contradicts the condition of the lemma, and the proof is complete.

Function definition $f$ (exhibitors)

We see (see Lemma 1 and Lemma 5) that for any

a > 0

$a> 0$ lots of

E_{a}

$E_a$ limited. This allows you to define the function

f : R^{+} r i g h t a r r o w R

$f: R ^ + \ rightarrow R$ by putting

f (a) = s u p E_{a}

$f (a) = \ sup E_a$ and

f (0) = 1

$f (0) = 1$ . For any non-empty subsets

A

$A$ ,

B

$B$ sets

R

$R$ real numbers set

A \ cdot B = \ {x: x = a \ cdot b

$A \ cdot B = \ {x: x = a \ cdot b$ where

a i n A, b i n B}

$a \ in A, b \ in B \}$ .

Lemma 7. If $A \ geq 0$ , $B \ geq 0$ non-empty bounded subsets $R$ then $\ sup (A \ cdot B) = \ sup A \ cdot \ sup B$ .

Evidence

Because

A c d o t B l e q s u p A c d o t s u p B

$A \ cdot B \ leq \ sup A \ cdot \ sup B$ then

s u p (A c d o t B) l e q s u p A c d o t s u p B

$\ sup (A \ cdot B) \ leq \ sup A \ cdot \ sup B$ . If a

s u p (A c d o t B) < s u p A c d o t s u p B

$\ sup (A \ cdot B) <\ sup A \ cdot \ sup B$ there is

v a r e p s i l o n > 0

$\ varepsilon> 0$ such that

s u p (A c d o t B) < t e x t s u p A c d o t t e x t s u p B - v a r e p s i l o n

$\ sup (A \ cdot B) <\ text {sup} A \ cdot \ text {sup} B- \ varepsilon$ . Therefore, for any

a i n A

$a \ in A$ and

b i n B

$b \ in B$ right

a b < t e x t s u p A (t e x t s u p B - v a r e p s i l o n)

$ab <\ text {sup} A (\ text {sup} B- \ varepsilon)$ (3)

Choose a sequence

\ left \ {a_n \ right \}

$\ left \ {a_n \ right \}$ elements of the set

A

$A$ converging to

s u p A

$\ sup A$ and sequence

\ left \ {b_n \ right \}

$\ left \ {b_n \ right \}$ elements of the set

B

$B$ converging to

s u p B

$\ sup B$ . But then

a_{n} b_{n} r i g h t a r r o w s u p A c d o t s u p B

$a_nb_n \ rightarrow \ sup A \ cdot \ sup B$ that contradicts

(3)

$(3)$ .

Lemma 7 is proved.

Lemma 8. For $a, b> 0$ fair equality $f (a + b) = f (a) \ cdot f (b)$ .

Evidence

Consider the sets

E_{a}

$E_a$ ,

E_{b}

$E_b$ and

E_{a + b}

$E_ {a + b}$ . Turning on

E_{a} c d o t E_{b} s u b s e t E_{a + b}

$E_a \ cdot E_b \ subset E_ {a + b}$ obviously. Let us prove that for any

z i n E_{a + b}

$z \ in E_ {a + b}$ there are

x i n E_{a}

$x \ in E_a$ and

y i n E_{b}

$y \ in E_b$ such that

x y g e q z

$xy \ geq z$ . Indeed let

z = l e f t (1 + a_{1} r i g h t) l e f t (1 + a_{2} r i g h t) l d o t s l e f t (1 + a_{k} r i g h t)

$z = \ left (1 + a_1 \ right) \ left (1 + a_2 \ right) \ ldots \ left (1 + a_k \ right)$ where

a_{1}, l d o t s, a_{k} > 0

$a_1, \ ldots, a_k> 0$ ,

a_{1} + l d o t s + a_{k} = a + b

$a_1 + \ ldots + a_k = a + b$ . Consider sets of positive numbers.

\ left \ {\ frac {a} {a + b} a_1, \ ldots, \ frac {a} {a + b} a_k \ right \}, \ left \ {\ frac {b} {a + b} a_1, \ ldots, \ frac {b} {a + b} a_k \ right \}

$\ left \ {\ frac {a} {a + b} a_1, \ ldots, \ frac {a} {a + b} a_k \ right \}, \ left \ {\ frac {b} {a + b} a_1, \ ldots, \ frac {b} {a + b} a_k \ right \}$ .

It's clear that

f r a c a a + b a_{1} + f r a c a a + b a_{2} + l d o t s + f r a c a a + b a_{k} = a

$\ frac {a} {a + b} a_1 + \ frac {a} {a + b} a_2 + \ ldots + \ frac {a} {a + b} a_k = a$ ,

f r a c b a + b a_{1} + f r a c b a + b a_{2} + l d o t s + f r a c b a + b a_{k} = b

$\ frac {b} {a + b} a_1 + \ frac {b} {a + b} a_2 + \ ldots + \ frac {b} {a + b} a_k = b$ .

Set

x = l e f t (1 + f r a c a a + b a_{1} r i g h t) l e f t (1 + f r a c a a + b a_{2} r i g h t) l d o t s l e f t (1 + f r a c a a + b a_{k} r i g h t)

$x = \ left (1+ \ frac {a} {a + b} a_1 \ right) \ left (1+ \ frac {a} {a + b} a_2 \ right) \ ldots \ left (1+ \ frac {a} {a + b} a_k \ right)$ ,

y = l e f t (1 + f r a c b a + b a_{1} r i g h t) l e f t (1 + f r a c b a + b a_{2} r i g h t) l d o t s l e f t (1 + f r a c b a + b a_{k} r i g h t)

$y = \ left (1+ \ frac {b} {a + b} a_1 \ right) \ left (1+ \ frac {b} {a + b} a_2 \ right) \ ldots \ left (1+ \ frac {b} {a + b} a_k \ right)$ .

It's clear that

x i n E_{a}

$x \ in E_a$ ,

y i n E_{b}

$y \ in E_b$ and

x c d o t y = l e f t (1 + f r a c a a + b a_{1} r i g h t) l e f t (1 + f r a c b a + b a_{1} r i g h t) l e f t (1 + f r a c a a + b a_{2} r i g h t) l e f t (1 + f r a c b a + b a_{2} r i g h t) l d o t s

$x \ cdot y = \ left (1+ \ frac {a} {a + b} a_1 \ right) \ left (1+ \ frac {b} {a + b} a_1 \ right) \ left (1+ \ frac {a} {a + b} a_2 \ right) \ left (1+ \ frac {b} {a + b} a_2 \ right) \ ldots$

l d o t s l e f t (1 + f r a c a a + b a_{k} r i g h t) l e f t (1 + f r a c b a + b a_{k} r i g h t) g e q l e f t (1 + a_{1} r i g h t) l e f t (1 + a_{2} r i g h t) l d o t s l e f t (1 + a_{k} r i g h t)

$\ ldots \ left (1+ \ frac {a} {a + b} a_k \ right) \ left (1+ \ frac {b} {a + b} a_k \ right) \ geq \ left (1 + a_1 \ right) \ left (1 + a_2 \ right) \ ldots \ left (1 + a_k \ right)$ ,

which completes the proof of Lemma 8.

so

s u p l e f t (E_{a} c d o t E_{b} r i g h t) = s u p E_{a + b}

$\ sup \ left (E_a \ cdot E_b \ right) = \ sup E_ {a + b}$ . But Lemma 7 implies that

s u p l e f t (E_{a} c d o t E_{b} r i g h t) = s u p E_{a} c d o t s u p E_{b}

$\ sup \ left (E_a \ cdot E_b \ right) = \ sup E_a \ cdot \ sup E_b$ .

We have built a valid function.

f

$f$ defined on a set of positive numbers such that

f (a + b) = f (a) c d o t f (b)

$f (a + b) = f (a) \ cdot f (b)$ . We finish it on the whole numerical line, putting

f (0) = 1

$f (0) = 1$ and

f (a) = f^{- 1} (- a)

$f (a) = f ^ {- 1} (- a)$ for any negative number

a

$a$ .

So function

f

$f$ defined on the whole number line.

Lemma 9. If $a + b = c$ then $f (a) \ cdot f (b) = f (c)$ .

Evidence

If one of the numbers

a

$a$ ,

b

$b$ ,

c

$c$ equally

0

$0$ then for them the statement of the lemma is true.

For the case when

a, b, c > 0

$a, b, c> 0$ Lemma follows from Lemma 8.

Further, if the lemma is true for numbers

a

$a$ ,

b

$b$ ,

c

$c$ then it is true for numbers

- a

$-a$ ,

- b

$-b$ ,

- c

$-c$ . Indeed, since

f (a) c d o t f (b) = f (c)

$f (a) \ cdot f (b) = f (c)$ then

f r a c 1 f (a) c d o t f r a c 1 f (b) = f r a c 1 f (c)

$\ frac {1} {f (a)} \ cdot \ frac {1} {f (b)} = \ frac {1} {f (c)}$ i.e.

f (- a) c d o t f (- b) = f (- c)

$f (-a) \ cdot f (-b) = f (-c)$ . So, it suffices to prove the lemma for the case

c > 0

$c> 0$ . But then either

a > 0

$a> 0$ ,

b > 0

$b> 0$ either

a > 0

$a> 0$ ,

b < 0

$b <0$ either

a < 0

$a <0$ ,

b > 0

$b> 0$ . Happening

a > 0

$a> 0$ ,

b > 0

$b> 0$ already disassembled. For definiteness, we set

a > 0

$a> 0$ ,

b < 0

$b <0$ . So,

a + b = c

$a + b = c$ , Consequently

a = c + (- b)

$a = c + (- b)$ where

a

$a$ ,

c

$c$ and

- b > 0

$-b> 0$ . So

f (a) = f (c) c d o t f (- b)

$f (a) = f (c) \ cdot f (-b)$ or

f (a) = f r a c f (c) f (b)

$f (a) = \ frac {f (c)} {f (b)}$ i.e.

f (a) c d o t f (b) = f (c)

$f (a) \ cdot f (b) = f (c)$ .

Lemma 9 is proved.

About function $f$

We built a function

f

$f$ defined on a set of real numbers such that for any

x, y i n R

$x, y \ in R$ right:

f (x) > 0

$f (x)> 0$ ,

f (x + y) = f (x) c d o t f (y)

$f (x + y) = f (x) \ cdot f (y)$ (four)

For

a > 0

$a> 0$ of

(2)

$(2)$ follows

f (a) g e q 1 + a

$f (a) \ geq 1 + a$ (five)

If

0 < a l e q f r a c 12

$0 <a \ leq \ frac {1} {2}$ then from

(2)

$(2)$ will get

f (a) l e q 1 + a + 2 a^{2}

$f (a) \ leq 1 + a + 2a ^ 2$ (6)

Note that because

0 < a l e q f r a c 12

$0 <a \ leq \ frac {1} {2}$ then

a + 2 a^{2} = a (1 + 2 a) l e q 2 a

$a + 2a ^ 2 = a (1 + 2a) \ leq 2a$ (7)

Finally out

(5)

$(5)$ ,

(6)

$(6)$ ,

(7)

$(7)$ will get

a l e q f (a) - 1 l e q a + 2 a^{2} l e q 2 a

$a \ leq f (a) -1 \ leq a + 2a ^ 2 \ leq 2a$ (eight)

It's clear that

f (y) - f (x) = f (x + (y x)) - f (x) = f (x) f (y x) - f (x) = f (x) (f (y x) - 1)

$f (y) -f (x) = f (x + (yx)) - f (x) = f (x) f (yx) -f (x) = f (x) (f (yx) -1)$ .

So, it is established that

f (y) - f (x) = f (x) (f (y - x) - 1)

$f (y) -f (x) = f (x) (f (y-x) -1)$ (9)

Estimate the value of

f (y - x) - 1

$f (y-x) -1$ . Putting in inequality

(8)

$(8)$

a = y - x

$a = y-x$ get for

x

$x$ ,

y

$y$ such that

x < y

$x <y$ and

y - x l e q f r a c 12

$y-x \ leq \ frac {1} {2}$ :

y - x l e q f (y - x) - 1 l e q (y - x) + 2 (y - x)^{2} l e q 2 (y - x)

$y-x \ leq f (y-x) -1 \ leq (y-x) +2 (y-x) ^ 2 \ leq 2 (y-x)$ (ten)

Using

(9)

$(9)$ from

(10)

$(10)$ we will receive:

f (x) (y x) l e q f (y) - f (x) l e q f (x) l e f t ((y x) + 2 (y x)^{2} r i g h t) l e q 2 f (x) (y x)

$f (x) (yx) \ leq f (y) -f (x) \ leq f (x) \ left ((yx) +2 (yx) ^ 2 \ right) \ leq 2f (x) (yx)$ (eleven)

T. to.

f (x) > 0

$f (x)> 0$ ,

(y - x) > 0

$(y-x)> 0$ then from

y > x

$y> x$ follows that

f (y) > f (x)

$f (y)> f (x)$ i.e.

f

$f$ increases by

R

$R$ . Further

0 < f (y) - f (x) l e q 2 f (x) (y - x)

$0 <f (y) -f (x) \ leq 2f (x) (y-x)$ that's why for

z > y > x

$z> y> x$ will get

| f (y) - f (x) | l e q 2 f (z) (y - x)

$| f (y) -f (x) | \ leq 2f (z) (y-x)$ (12)

Of

(12)

$(12)$ it follows that on the set

(- i n f t y; z]

$(- \ infty; z]$ function

f

$f$ uniformly continuous. So

f

$f$ continuous everywhere on

R

$R$ .

Now we estimate the value of the derivative of the function

f

$f$ at an arbitrary point

x i n R

$x \ in R$ .

Let be

x_{n} < y_{n}

$x_n <y_n$ and

x_{n} r i g h t a r r o w x

$x_n \ rightarrow x$ ,

y_{n} r i g h t a r r o w x

$y_n \ rightarrow x$ at

n r i g h t a r r o w i n f t y

$n \ rightarrow \ infty$ . Then

f r a c f l e f t (x_{n} r i g h t) l e f t (y_{n} - x_{n} r i g h t) y_{n} - x_{n} l e q f r a c f l e f t (y_{n} r i g h t) - f l e f t (x_{n} r i g h t) y_{n} - x_{n} l e q f r a c f l e f t (x_{n} r i g h t) l e f t (y_{n} - x_{n} + 2 l e f t (y_{n} - x_{n} r i g h t)^{2} r i g h t) y_{n} - x_{n}

$\ frac {f \ left (x_n \ right) \ left (y_n-x_n \ right)} {y_n-x_n} \ leq \ frac {f \ left (y_n \ right) -f \ left (x_n \ right)} {y_n-x_n} \ leq \ frac {f \ left (x_n \ right) \ left (y_n-x_n + 2 \ left (y_n-x_n \ right) {} ^ 2 \ right)} {y_n-x_n}$ ,

i.e.

f l e f t (x_{n} r i g h t) l e q f r a c f l e f t (y_{n} r i g h t) - f l e f t (x_{n} r i g h t) y_{n} - x_{n} l e q f l e f t (x_{n} r i g h t) l e f t (1 + 2 l e f t (y_{n} - x_{n} r i g h t) r i g h t)

$f \ left (x_n \ right) \ leq \ frac {f \ left (y_n \ right) -f \ left (x_n \ right)} {y_n-x_n} \ leq f \ left (x_n \ right) \ left ( 1 + 2 \ left (y_n-x_n \ right) \ right)$ .

Because

f l e f t (x_{n} r i g h t) r i g h t a r r o w f (x)

$f \ left (x_n \ right) \ rightarrow f (x)$ at

n r i g h t a r r o w i n f t y

$n \ rightarrow \ infty$ and

f l e f t (x_{n} r i g h t) l e f t (1 + 2 l e f t (y_{n} - x_{n} r i g h t) r i g h t) r i g h t a r r o w f (x)

$f \ left (x_n \ right) \ left (1 + 2 \ left (y_n-x_n \ right) \ right) \ rightarrow f (x)$ at

n r i g h t a r r o w i n f t y

$n \ rightarrow \ infty$ then

f r a c f l e f t (y_{n} r i g h t) - f l e f t (x_{n} r i g h t) y_{n} - x_{n} r i g h t a r r o w f (x)

$\ frac {f \ left (y_n \ right) -f \ left (x_n \ right)} {y_n-x_n} \ rightarrow f (x)$ .

It means that

f

$f$ everywhere differentiable on

R

$R$ and

f^{'} (x) = f (x)

$f '(x) = f (x)$ .

Slobodnik Semen Grigorievich ,
content developer for the application "Tutor: Mathematics" (see the article on Habré ), Ph.D. in Physics and Mathematics, teacher of school mathematics 179 Moscow

Source: https://habr.com/ru/post/416863/

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