Exercises in Maths III
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Transcript of Exercises in Maths III
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21 - 30
, 2014
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21(-(30
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, . http://lisari.blogspot.gr/2012/04/25.html
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21 f : [,] ! , [,] , f (x) 0 ,
x [,] , z, Re(z) 0 , Im(z) 0 , | Re(z) | > | Im(z) | .
z +
1z
= f () z 2 +
1z 2
= f 2() , :
) | z | = 1 . ) f2() < f 2() .
) x3 f ()+ f () = 0 , , (1,1) .
) z +
1z
= f () ! , z +
1z
= z +1z
, '
z2z + z = z 2z + z z 2z z 2z + z z = 0 zz (z z)(z z) = 0 (z z)(zz 1) = 0
z z = 0 zz 1 = 02 i Im(z) = 0 | z |2= 1
Im(z) = 0 (, Im(z) 0 ) | z | = 1 | z | = 1 .
) z +
1z
= f ()
z +
1z
2
= f 2() z 2 +1z 2
+ 2 = f 2() f 2()+ 2 = f 2() f 2() f 2() = 2 < 0
f 2() f 2() < 0 f 2() < f 2() (1)
) g(x) = x3 f ()+ f () , x [1,1] ,
g(x) = 0 , , (1,1) .
. g [1,1] .
. g(1) = (1)3 f ()+ f () g(1) = f () f () .
. g(1) = 13 f ()+ f () g(1) = f ()+ f () .
g(1) g(1) = [f () f ()][f ()+ f ()] = f2() f 2() < 0 , (1).
Bolzano, g(x) = 0 , , (1,1) .
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22 f : [1,+) ! , f (x) = x
2 nx .
) f.
) f (x) = 2012 .
) z1 ,z2 ! , | z1 | >1 , | z2 | >1 z1z1 n | z1 | = z2z2 n | z2 | = n16 , :
. | z1 | = | z2 | = 2 . .
z1
+ z2
4 + z1z
2
! .
) zz n | z | = 1 , z , > 1.
) f = [1,+) -, , ,
f (x) = (x 2 ) nx + x 2 (nx ) = 2x nx + x 2
1x
= 2x nx + x f (x) = x(2nx +1) > 0 ,
x >1 , nx > 0 , x >1 .
f [1,+) , .
) x 1 , nx 0 , f (x) 0 , f () = [0,+) .
2012 f () , f (x) = 2012 , , .
, , f , .
f , f () = [ f (1) , im
x+f (x)) .
f (1) = 12 n1 = 1 0 f (1) = 0 .
imx+
f (x) = imx+
(x 2 nx) = (+) (+) = + .
f () = [0,+) .
2012 f () , f (x) = 2012 , , .
, , f , .
) . z1z1 n | z1 | = z2z2 n | z2 | = n16
| z1 |2 n | z
1| = | z
2|2 n | z
2| = n24 f (| z
1|) = f (| z
2|) = 4n2
f (| z1 |) = f (| z2 |) = 22 n2 f (| z
1|) = f (| z
2|) = f (2) .
f , 1-1, -
| z
1| = | z
2| = 2 .
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. w =
z1
+ z2
4 + z1z
2
, w = w .
w =
z1
+ z2
4 + z1z
2
.
| z1 | = 2 | z
1|2 = 4 z
1z
1= 4 z
1=
4z
2
. z
2=
4z
2
.
w =
4z
1
+4z
2
4 +4z
1
4z
2
=
41z
1
+1z
2
4 1+4
z1z
2
=
1z
1
+1z
2
1+4
z1z
2
=
z2
+ z1
z1z
2
z1z
2+ 4
z1z
2
=z
1+ z
2
4 + z1z
2
= w
w = w .
) zz n | z | = 1 | z |2 n | z | = 1 f (| z |) = 1 (1)
f (x) > 0 , x >1 , 1 = (1,+) , f (1) = (0,+) .
1 f () , >1 , f () = 1 .
f 1 , .
, (1) f (| z |) = f () f 11
| z | = , >1 , z
(0, 0) > 1.
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23 z = + i , , ,
f :! ! ,
f (x) = | t z + z |dt
1
x2+1
3x + 2 . f - x0 = 1 , :
) | z | =
12
.
) w = 2z i - .) w, .
) , f, x'x x A 0 x A 1.
) | tz + z | = | z(t +1) | = | z | | t +1 | ,
f (x) = | z | | t +1 |dt
1
x2+1
3x + 2 f (x) = | z | | t +1 |dt1
x2+1
3x + 2 . | t +1 | ! ,
| t +1 |dt
1
x
! .
,
| t +1 |dt
1
x2+1
! x 2 +1
| t +1 |dt
1
x
, f ! -, 1, .
, , 1 D
f= ! , Fermat
f (1) = 0 (1)
f (x) = | z | | x2 +1+1 | (x 2 +1 ) 3 = | z | | x 2 + 2 | 2x 3
f (x) = 2x | z | | x2 + 2 |3 .
(1) 2 1 | z | | 12 + 2 |3 = 0 6 | z | = 3 | z | =
36 | z | =
12
.
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) w = 2z i 2z = w + i z =
w + i2
.
| z | =
12
w + i2
=12
|w + i |2
=12 |w + i | = 1 ,
(0,1) 8 1.
) w 2i , - .
) ,
[0,1] , f ' .
| z | =
12
(), f (x) =
12 | x 2 + 2 | 2x 3 .
x2 + 2 > 0 , x ! , | x
2 + 2 | = x 2 + 2 ,
f (x) = x(x2 + 2)3 = x 3 + 2x 3 f (x) = (x 1)(x 2 + x + 3) .
12 4 1 3 = 11 < 0 , x
2 + x + 3 > 0 ,
x ! , ., f (x) > 0 x 1 > 0 x >1 .
f x 8 1, -
f (x) f (1) (2) , x ! .
f (1) =12
| t +1 |dt
1
12+1
3 1+ 2 f (1) = 12 | t +1 |dt1
2
1 . t +1 > 0 , t [1,2] , | t +1 | = t +1 ,
| t +1 |dt
1
2
= (t +1)dt1
2
= t2
2+ t
1
2
=22
2+ 2
12
2+1
=52
.
f (1) =
12521 =
541 f (1) =
14
.
, (2) f (x)
14
, x ! , -
= f (x)dx
0
1
= 1 f (x)dx0
1
= (x ) f (x)dx0
1
= [x f (x)]01 x f (x)dx0
1
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1 1 +
f (x) +
f (x) 2 1
= 1 f (1)0 f (0) x(x 3 + 2x 3)dx
0
1
= 14 (x 4 + 2x 2 3x)dx0
1
=
14
x 5
5+
2x 3
3
3x 2
2
0
1
.
g(x) =
x 5
5+
2x 3
3
3x 2
2,
=
14[g(x)]
01 =
14[g(1)g(0)] =
14g(1)+ g(0) (3)
g(1) =
15
5+
2 13
3
3 12
2=
15
+23
32
=6 + 20 45
30 g(1) =
1930
.
g(0) =
05
5+
2 03
3
3 02
2 g(0) = 0 .
(3) =
141930
+ 0 =14
+1930
=15 + 38
60 =
5360
.. .
24 z = + i (, !) f (x) = | z xi | , x ! , (1,2) .
) 1 | z | 3 . z ;
) f (x) = 2 | z | x , , (1,1) .
) C
f (1,2) , f (1) = 2 , '
| z 1 i | = 2 | z i | = 2 ,
(0,1) D 2.
,
() () () 1 | z | 3 .
| z | = 1 , ,
z =i .
, | z | = 3 , ,
z = 3i .
) f (x)2 | z | x = 0 .
g(x) = f (x)2 | z | x , x [1,1] , g(x) = 0
, , (1,1) .
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. f (x) = | z xi | = |+ ixi | = |+ (x)i | f (x) = 2 + (x)2 .
f ! , g [1,1] .
. g(1) = f (1)2 | z | (1) = | z (1) i | + 2 | z | g(1) = | z + i | + 2 | z | .
| | z | | i | | | z + i | | z | + | i | | | z |1 | | z + i | | z | +1 .
() | z |1 | z |1 0 , | | z |1 | = | z |1 .
| z |1 | z + i | 2 | z | + | z |1 | z + i | + 2 | z | 3 | z |1 g(1) .
| z |1
3 | z | 3 3 | z |1 31 3 | z |1 2 > 0 g(1) > 0 .
III. g(1) = f (1)2 | z | 1 = 22 | z | =2(| z |1) g(1) 0 , | z |1 0 .
g(1) g(1) 0 , :
i) g(1) g(1) < 0 , Bolzano g(x) B 0
, , (1,1) .
ii) g(1) g(1) = 0 , g(1) = 0 ( g(1) > 0 ), 1
g(x) B 0.
(i) (ii) , , g(x) = 0 , -
, [1,1) .
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25 f : [,] ! z = + i , z1 = + f () i
z2 = + f () i . 3(z2 z 2) 4izz = 4 i Re(z
1z
2) ,
C
f ,
, x'x. f (x) = 0 , , (,) .
z2 z 2 = (z z )(z + z ) = 2 i Im(z) 2Re(z) = 2i 2 = 4 i .
, z1z2 = [+ f () i ][ f () i ] = f () i + f () i f () f () i2
z1z2 = [ + f () f ()]+ [ f () f ()]i .
,
3 4 i 4i | z |2 = 4i [ + f () f ()] 3(2 + 2) = + f () f ()
f () f () = 22 2 f () f () =(2 + 2 2) f () f () =()2 .
[,] < < 0 ()2 > 0 ,
f () f () < 0 .
f [,] , Bolzano,
f (x) = 0 , , (,) .
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26) z. z (1+ i)z + (1 i)z + 4 = 0 , z, , ().
) () () -
f, f (x) = x + +
x 2
ex + , , ! .
) g(x) =
f (x)x
, x 0 . - g.
)
z + iz + z i z + 4 = 0 (z + z )+ i(z z )+ 4 = 0 2Re(z)+ i 2 i Im(z)+ 4
