szeregi potegowe- odpowiedzi.pdf

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Szeregipot¦gowe:odpowiedzi,rozwi¡zania,wskazówki
Uwaga: Wewszystkichzadaniach D oznaczazbiórtych x 2 R ,dlaktórychrozwa»anyszeregpot¦gowy
jestzbie»ny.
1.1 D =[ 5 , 5] 1.2 D =[ 2 , 0] 1.3 D =( 3 , 3] 1.4 D =[0 , 2) 1.5 D = 1 3 , 1 3
1.6 D = 3 2 , 5 2
1.7 D =(0 , 1](Wsk.mo»eszpodstawi¢ t =1 2 x )
1.8 D =( 8 , 0) 1 .9 D = { 0 } 1.10 D = 9 2 , 3 2 1.11 D =( 3 , 3)
7) 1.13 D =[ 7 , 9)(Wsk.mo»eszpodstawi¢ t =( x 8) 3 )
1.14 D =[ 2 , 2) 1.15 D =
7 , 5+
p
2 p 3 4
3 , 2 p 3 4
i
(Wsk.mo»eszpodstawi¢ t =(3 x +4) 2 )
3
x 0 = 2, R = 2 3 , D = 8 3 , 4 3
Zadanie3
3.1 D =( 1 , 1)
1 P
( 1) n x 2 n +1 0
= x
1+ x 2
0
P
( 1) n (2 n +1) x 2 n =
P
( 1) n x 2 n +1 0 =
= 1 x 2
(1+ x 2 ) 2
S ( x )=
n =0
n =0
n =0
albo
S ( x )=
x R
0
1 P
( 1) n x R
0
0
1 P
( 1) n x 2 n +1 0
x
1+ x 2
0
S ( t ) dt
=
(2 n +1) t 2 n dt
=
=
= 1 x 2
(1+ x 2 ) 2
n =0
n =0
0
3.2 D =( 1 , 1]
S ( x )=
1 P
( 1) n x 2 n +1
2 n +1 =
1 P
( 1) n x R
0
t 2 n dt =
x R
1 P
( 1) n t 2 n dt =
x R
1 P
( t 2 ) n dt =
x R
1+ t 2 dt =arctg x
1
n =0
n =0
0
n =0
0
n =0
0
albo
S 0 ( x )=
1 P
( 1) n x 2 n =
1 P
( x 2 ) n = 1
1+ x 2 ,awtedy S ( x )=
x R
S 0 ( t ) dt + S (0)=
x R
1+ t 2 dt +0=arctg x
1
3.3 D = 4 3 , 4 3 , S ( x )=ln 4
n =0
n =0
0
0
4 3 t (Wsk.rozwi¡zujesi¦jakzadanie3.2)
3.4 D =( 1 , 1), S ( x )= 1
(1 x ) 2 (Wsk.rozwi¡zujesi¦jakzadanie3.1)
3.5 D =( 1 , 1)
1 P
x n +1 00
= x 2
1 x
00
P
n ( n +1) x n 1 =
P
1
(( n +1) x n ) 0 =
P
x n +1 00 =
= 2
(1 x ) 3
S ( x )=
3.6 D = 4 5 , 4 5 , S (0)= 1 2 ,za±dla x 6 =0mamy
S ( x )=
n =1
n =1
n =1
n =1
1 P
4 n +1 ( n +2) x n = 1 x
5 n
1 P
4 n +1 ( n +2) x n +1 = 1 x
5 n
1 P
4 n +1 x n +2 0 ,adalej,jakwzadaniu3.1.
5 n
n =0
n =0
n =0
Otrzymujemy S ( x )= 8 5 x
(4 5 x ) 2
3.7 D =
p
p
3
3
3 ,
3
S ( x )=
1 P
2 n +1 x 2 n +1 =
3 n
1 P
3 n x R
0
t 2 n dt ,adalejjakwzadaniu3.2.Mo»nate»policzy¢najpierw S 0 ( x )=
n =0
n =0
x R
x R
3 ln 1+ p 3 x
... = 1
1 3 x 2 ,awtedy S ( x )=
S 0 ( t ) dt + S (0)=
1 3 t 2 dt +0= 1
1
p
p
2
1
3 x
3.8 D = 2 p 5
, S (0)=3,za±dla x 6 =0mamy
0
0
5 , 2 p 5
5
1 P
1 P
5 4 n (2 n +3) x 2 n +2 = 1 x 2
1 P
5 4 n x 2 n +3 0 ,adalej,jakwzadaniu
S ( x )=
4 n (2 n +3) x 2 n = 1 x 2
n =0
n =0
n =0
(4 5 x 2 ) 2 .
Mo»natak»erozwa»y¢funkcj¦pomocnicz¡ f ( x )= x 2 S ( x )=
P
1
4 n (2 n +3) x 2 n +2 ,awtedy f ( x )=
x R
0
1 P
5 4 n x 2 n +3 0
n =0
= 4 x 3
0
f ( t ) dt
=
= 48 x 2 20 x 4
(4 5 x 2 ) 2 .Zatem S ( x )= 1 x 2 f ( x ).
4 5 x 2
0
n =0
1 P
n t n
x R
x R
P
P
3.9 D =[ 1 , 1], S ( x )= 1 x
n ( n +1) x n +1 = 1 x
( 1) n
n
t n dt = 1 x
( 1) n
dt .Dalejsum¦
n =1
n =1
0
0
n =1
podcałk¡liczymyjeszczeraztak,jakwzadaniu3.2.Otrzymujemy S ( x )=1 (1+ x )ln(1+ x )
x dla x 6 =0i
S (0)=0.(Uwaga:sprawd¹,»e S jestci¡gławzerze!).
1
Zadanie1
1.12 D =(5 p
Zadanie2
1
1
1
1
5 n
3.1.Otrzymujemy S ( x )= 48 20 x 2
5 n
1
1
( 1) n
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Zadanie4
Wka»dymprzykładziewprowadzonofunkcj¦ S ( x )tak,abyszukanasumaszeregubyłarówna S (1).Sum¦
S ( x )mo»nawyznaczy¢metodami,opisanymiwpoprzednimzadaniu.
4.1 D =( 4 , 4), S ( x )=
P
4 n x n = 16
(4 x ) 2 ,st¡d S (1)=
P
4 n = 16 9
n =0
n =0
4.2 D = 5 3 , 5 3 S ( x )=
P
3 5 n x n =ln 5
P
3 5 n =ln 5 8
1
1
( 1) n
n
5+3 x ,st¡d S (1)=
( 1) n
n
n =1
n =1
1 P
1 P
4.3 D =( 3 , 3), S ( x )=
9 n (2 n +1) x 2 n +1 = 81 9 x 2
(9+ x 2 ) 2 ,st¡d S (1)=
9 n (2 n +1)= 72
100
n =0
n =0
P
P
1
1
4.4 D =( 5 , 5), S ( x )=
5 n x n +1 = 50 x 5 x 2
(5 x ) 2 ,st¡d S (1)=
5 n = 45 16
n =0
n =0
1 P
1 P
4.5 D =[ 7 , 7), S ( x )=
n 7 n x n =ln 7
7 x ,st¡d S (1)=
n 7 n =ln 7 6
n =1
n =1
1 P
1 P
1 P
4.6 D =( 4 , 4), S ( x )=
( 4) n +1 x n 1 = 32 · 16
(16+4 x ) 3 ,st¡d S (1)=
n ( n +1)
( 4) n +1 =
( 4) n +1 = 8 125
n =1
n =0
n =1
2
1
1
n +1
n +1
( 1) n
( 1) n
n +2
n +2
1
1
n ( n +1)
n ( n +1)
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