Uzdevumi 15 2 sem


R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
15. nodarb%2łba
1
1. piemrs. Apr7int x sin x dx ar precizitti l%2łdz  = 0,0001.
+"
0
1 1 1
n
" "
# ś#
(
x sin x dx = x
"(-1)n x2n+1 ź#dx "(2-1)
+" +"ś# n=0 (2n +1)!ź# = n=0 n +1)!0 x " x2n+1dx =
+"
ś#
# #
0 0
1
5
2n+
3
1
n n n
" " "
2n+ 2
( ( (
=
"(2-1) "(2-1) x "(2-1) 1
+"x 2 dx = n=0 n +1)!(2n + 2,5) = n=0 n +1)!(2n + 2,5) H"
n +1)!0
n=0
0
1 1 1 1
H" - + - H" 0,36408.
2,5 6" 4,5 120"6,5 720"8,5
0,5
2. piemrs. Apr7int (1+ x3)dx ar precizitti l%2łdz  = 0,0001.
+"ln
0
0,5 0,5
"
#
n
)n+1 ś# " (-1)n+1 0,5x3ndx =
ś# ź#dx
(1+ (x3) =
"(-1n "
+"ln x3)dx = +" +"
ś#n=1 ź#
n
n=1
# #
0 0 0
0,5
"
)n+1 x3n+1 " (-1)n+1 3n+1 1 1 1
#
= H" - + H" 0,0151.
ź#
"(-1n (3n +1) = n=1n(3n +1)"ś# 1 ś#
"
2
# # 4 " 24 14 " 27 30 " 210
n=1
0
1
1- cos x
3. piemrs. Apr7int integrli dx ar precizitti l%2łdz 10-3 (izmantojot
+"
x2
0
zemintegr<a funkcijas izvirz%2łaanu rind).
Zemintegr<a funkciju izvirz%2łsim pakpju rind, izmantojot funkcijas cos x izvirz%2łjumu
n
"
(-1) x2n
cos x = :
"
(2n)!
n=0
"
(-1)n ź#
(-1)n x2n 1- # " x2n ś#
ś#1+
1-
"
"
ś# ź#
"
(2n)!
(2n)!
1- cos x (-1)n x2n " (-1)n+1 x2n-2
n=1
# #
n=0
= = = - =
" "
(2n)!
x2 x2 x2 (2n)!"x2 n=1
n=1
15. nodarb%2łba. 1. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
Tad
1 1
"
#
1- cos x (-1)n+1 x2n-2 ś# " (-1)n+1 1 x2n-2dx =
dx =
" "
+" +"ś# n=1 (2n)! ź# dx = n=1 (2n)! +"
ś# ź#
x2
# #
0 0 0
1
"
(-1)n+1 x2n-1 " (-1)n+1 1 1 1
= = H" - + H" 0,486 .
" "
(2n)! 2n -1 (2n)!(2n -1) 2 72 3600
n=1 n=1
0
x
2
4. piemrs. Atrast integrli F(x)= dt un apr7int F(0,2) ar precizitti 10-4.
+"e-t
0
`is integrlis k nenoteiktais integrlis nav izsakms caur elementrajm
funkcijm gal%2łg veid (tas ir elementri neintegrjams). Tpc izmantosim rindu
"
xn
ex = ,
"
n!
n=0
kura konver# visiem xS (-ś; ś).
2
Ievietojot x viet - t , iegksim
n
" "
2
(- t2) (-1)nt2n
e-t = = .
" "
n! n!
n=0 n=0
Iegkto rindu nointegrjot robe~s no 0 l%2łdz x, varam rakst%2łt
x
x x x
n
" " "
# ś#
2
1 (-1)nt2n+1
F(x) = dt =
"ś# (-1)nt2n dt ź# = "(-n!) "
+"e-t n=0ś#+" +"t2ndt = n=0 n!(2n +1) =
ź#
n!
n=0
0 # 0 # 0
0
"
(-1)n x2n+1
= .
"
n!(2n +1)
n=0
Tagad atrad%2łsim
"
1 (-1)n = 1 1 1
F(0,2) = F# ś# = - + - ... .
ś# ź#
"
5 5
# # 52n+1n!(2n +1) 53 " 3 55 " 2 " 5
n=0
Esam ieguvuai alternjoau rindu (ir jievro, ka 0!= 1). T k
1
H" 0,000032 < 10-4 , tad pdjo saskaitmo var neFemt vr un
55 " 2 " 5
15. nodarb%2łba. 2. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
1 1 1 1 74
F(0,2) H" - = - = = 0.1996 .
5 5 375 375
53 "3
x
2
2
Piez%2łme. Funkciju erf (x)=
+"e-t dt sauc par k<kdas funkciju. Funkciju
Ą
0
x
2
2
Ś(x)=
+"e-t dt sauc par varbkt%2łbu integrli vai Laplasa funkciju. `%2łs funkcijas
2Ą
0
plaai lieto varbkt%2łbu teorij.
5. piemrs. Ar pakpju rindas pal%2łdz%2łbu, Femot pirmos seaus izvirz%2łjuma locek<us,
2
atrast diferencilviendojuma y = x2 + y2 partikulro atrisinjumu pie skuma
nosac%2łjuma y(0) = 1.
Diferencilviendojuma atrisinjumu meklsim Maklorena rindas form, t.i.,
(5)
2 2
f (0) f (0)
2
f (x) H" f (0)+ f (0)x + x2 + ... + x5 .
2! n!
Noteiksim atvasinjumu (l%2łdz piektajai pakpei ieskaitot) vrt%2łbas punkt x = 0:
2 2
f (0)= y (0)= 02 + y(0)2 = 0 +1 =1.
2 2 2 2
Lai atrastu f (0)= y (0), atvasinsim diferencilviendojuma abas puses, t iegkstot:
2 2 2 2 2 2 2
y = 2x + 2yy , f (0) = y (0) = 2 " 0 + 2 "1"1 = 2 .
L%2łdz%2łgi atrad%2łsim prjos koeficientus:
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
y = 2 + 2y " y + 2yy = 2 + 2(y )2 + 2yy , f (0) = y (0) = 2 + 2 "12 + 2 "1" 2 = 8 ;
(4)
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
y(4) = 4y y + 2y y + 2yy = 6y y + 2yy , f (0) = y(4)(0) = 6 "1" 2 + 2 "1"8 = 28 ;
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
y(5) = 6y " y + 6y " y + 2y y + 2yy(4) = 6(y )2 + 8y " y + 2yy(4),
(5)
f (0) = y(5)(0) = 6 " 22 + 8 "1"8 + 2 "1" 28 = 24 + 64 + 56 = 144 .
Tagad varam rakst%2łt aptuveno diferencilviendojuma atrisinjumu, Maklorena rind
ievietojot apr7intos skait<us:
x 2x2 8x3 28x4 144x5 4x3 7x4 6x5
y H" 1+ + + + + = 1+ x + x2 + + + .
1! 2! 3! 4! 5! 3 6 5
15. nodarb%2łba. 3. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
6. piemrs. Tuvinti ar rindu pal%2łdz%2łbu atrisint diferencilviendojumu
2
y = y3 - x, y(2)=1 (atrast pirmos 4 nenulles atrisinjuma locek<us).
2
No dot y(2)= 1; y (2)=13 - 2 = -1;
2 2 2 2 2
y = 3y2 y -1 , y (2) = 3"12 "(-1)-1 = -4;
2 2 2 2 2 2 2 2 2
y = 6y(y )2 + 3y2 y , y (2) = 6 "1"(-1)2 + 3"12 "(- 4) = -6 .
Dot diferencilviendojuma tuvints atrisinjums ir
-1 - 4 - 6
y H" 1+ (x - 2)+ (x - 2)2 + (x - 2)3 = 1- (x - 2)- (x - 2)2 - (x - 2)3.
1! 2! 3!
2 2 2
7. piemrs. Atrisint diferencilviendojumu y ln x - sin xy = 0 , y(e) = 0 , y (e) = e-1
ar rindu pal%2łdz%2łbu (atrast 5 atrisinjuma locek<us).
Dot diferencilviendojuma atrisinjumu izvirz%2łsim pakpju rind pc (x - e)
pakpm, td< noteiksim atrisinjuma atvasinjumu vrt%2łbas punkt x = e :
2 2 2 2 2 2
y (e)ln e - sin(e " 0) = 0 ! y (e)"1- 0 = 0 ! y (e) = 0 .
Atvasinsim doto viendojumu
2 2
y
2 2 2 2
y ln x + - cos xy "(y + xy ) = 0 .
x
Tad
0
2 2 2 2 2 2 2 2 2
y (e)ln e + - cos(e " 0)"(0 + e " e-1)= 0 ! y (e)"1+ 0 -1"(0 +1) = 0 ! y (e) = 1
e
Vlreiz atvasinot, iegkstam
2 2 2 2 2 2 2 2
y y x - y
2 2 2 2 2
yIV ln x + + + sin xy "(y + xy )2 - cosxy "(y + y + xy ) = 0 ,
x
x2
2
1 1" e - 0
yIV (e)lne + + + sin(e " 0)(0 + e " e-1) - cos(e " 0)"(e-1 + e-1 + e " 0)= 0 !
e
e2
1 1 2
IV IV
y (e)"1+ + + 0 -1" = 0 ! y (e) = 0 .
e e e
Ttad
e-1 0 1 0 1 1
y H" 0 + (x - e)+ (x - e)2 + (x - e)3 + (x - e)4 = (x - e)+ (x - e)3 .
1! 2! 3! 4! e 6
15. nodarb%2łba. 4. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
8. piemrs. Iegkt rindu skait<a Ą apr7inaanai, izmantojot funkciju arctg x.
Izvirz%2łsim funkciju y = arctg x pakpju rind:
"
x2n-1
n-1
arctg x =
"(-1) 2n -1 .
n=1
Td gad%2łjum
"
Ą
n-1
arctg 1 = =
"(-1) 2n1 .
4 -1
n=1
Sekojoai esam ieguvuai
"
4
n-1
Ą = 4
"(-1) 2n1 = 4 - 4 + 5 - 4 + ... + (-1)n-1 4 + ... .
-1 3 7 2n -1
n=1
Tagad pieFemsim, ka skaitlis Ą japr7ina ar precizitti 0,001. Td gad%2łjum ir
4 1
jizpilds neviend%2łbai d" jeb 2n -1 e" 4000 . L%2łdz ar to ir jizpilds
2n -1 1000
2n e" 4001 fl n e" 2000,5 . Tas noz%2łm, ka, lai sasniegtu pras%2łto precizitti, ir jFem
2001 saskaitmais.
Labkus rezulttus iegkstam, ja funkcijas y = arctg x izvirz%2łjuma rind Femsim
1
vrt%2łbu x = . Td gad%2łjum
3
"
1 Ą 1
n-1
arctg = =
"(-1)
2n-1
6
3
n=1 ( 3) (2n -1)
un
"
6
n-1
Ą = .
"(-1)
2n-1
n=1 ( 3) (2n -1)
6 6
< 0,001, ja n e" 7 (pie n = 7 iegksim a7 = H" 0,000366 ).
2n-1 13
( 3) (2n -1) ( 3) "13
Ttad, lai apr7intu skaitli Ą ar doto precizitti, pietiek Femt 6 rindas locek<us.
Tdjdi
" # ś#
1 1 1 1 1
n-1
Ą = 6 = 6ś# - + - + ...ź# H"
"(-1)
2n-1 3 5 7
ś# ź#
3
n=1 ( 3) (2n -1) ( 3) " 3 ( 3) " 5 ( 3) " 7
# #
# ś#
1 1 1 1 1 1
ś#
H" 2 3 " 3ś# - + - + - ź# =
3 3 3 " 3
32 3 " 5 33 3 " 7 34 3 " 9 35 3 "11ź#
# #
1 1 1 1 1
ś#
= 2 3#1 - + - + - ź#
H" 3,14131.
ś#
9 45 189 729 2673
# #
15. nodarb%2łba. 5. lpp. Augstk matemtika.
I. Volodko


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