Лабораторные работы по численным методам. Пак Т.В. - 21 стр.

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Составители: 

x
n
= x
n1
f(x
n1
)(¯x
n1
x
n1
)
f(¯x
n1
) f(x
n1
)
, n = 0, 1, 2, . . . ,
¯x
0
, ¯x
1
, . . .
x
0
, x
1
, . . .
|x
n
¯x
n
)| < ε
0 = 1.2x
2
sin(10x)
0 = 2
x cos(πx/2)
0 = 2
x
2x
2
1
0 = 2 ln x 1/x
0 = 2 lg x x/2 + 1
0 = lg x 7/(2x + 6)
0 = x lg x 1/2
0 = lg(3x 1) + exp(2x 1)
0 = exp(x) 2(x 1)
2
0 = 2 x exp(x)
0 = 1/x π cos(πx)
0 = sec(x) x
2
1
0 = ctg(1, 05x) x
2
0 = 2x lg(x) 7
0 = exp(x) + x
2
2
0 = 0, 5x
2
cos(2x)
0 = ln(0, 5x) 0, 5 cos(x)
0 = ln(2x) exp(2x)
0 = exp(x) + x
3
3
0 = 2x
2
cos(2x)
0 = x
2
20 sin x
0 = x
2
sin 5x
0 = ln x + (x + 1)
3
0 = 2, 2x 2
x
                          f (xn−1 )(x̄n−1 − xn−1 )
            xn = xn−1 −                            , n = 0, 1, 2, . . . ,
                            f (x̄n−1 ) − f (xn−1 )
x̄0 , x̄1 , . . .  ïðèáëèæåíèÿ ïî ìåòîäó êàñàòåëüíûõ,
x0 , x1 , . . .  ïðèáëèæåíèÿ ïî ìåòîäó õîðä.
      Âû÷èñëåíèÿ ïî ôîðìóëàì ïðåêðàùàþò, êîãäà |xn − x̄n )| < ε.
      Òàáëèöà 5.

      Âèä óðàâíåíèÿ                   èñêîìûé êîðåíü
       f(x)=0
 1     0 = 1.2x2 − sin(10x)            âñå ïîëîæèòåëüíûå êîðíè
            √
 2     0 = 2 x − cos(πx/2)             âñå êîðíè
 3     0 = 2x − 2x2 − 1                ïîëîæèòåëüíûå êîðíè
 4     0 = 2 ln x − 1/x                âñå êîðíè
 5     0 = 2 lg x − x/2 + 1            ïîëîæèòåëüíûå êîðíè
 6     0 = lg x − 7/(2x + 6)           âñå êîðíè
 7     0 = x lg x − 1/2                âñå êîðíè
 8     0 = lg(3x − 1) + exp(2x − 1)    âñå êîðíè
 9     0 = exp(−x) − 2(x − 1)2         âñå êîðíè
 10    0 = 2 − x exp(x)                âñå êîðíè
 11    0 = 1/x − π cos(πx)             âñå ïîëîæèòåëüíûå êîðíè
 12    0 = sec(x) − x2 − 1             âñå ïîëîæèòåëüíûå êîðíè
 13    0 = ctg(1, 05x) − x2            âñå ïîëîæèòåëüíûå êîðíè
 14    0 = 2x − lg(x) − 7              âñå ïîëîæèòåëüíûå êîðíè
 15    0 = exp(−x) + x2 − 2            îòðèöàòåëüíûå êîðíè
 16    0 = 0, 5x2 − cos(2x)            âñå êîðíè
 17    0 = ln(0, 5x) − 0, 5 cos(x)     âñå êîðíè
 18    0 = ln(2x) − exp(2x)            âñå êîðíè
 19    0 = exp(−x) + x3 − 3            âñå ïîëîæèòåëüíûå êîðíè
 20    0 = 2x2 − cos(2x)               âñå êîðíè
 21    0 = x2 − 20 sin x               âñå êîðíè
 22    0 = x2 − sin 5x                 âñå êîðíè
 23    0 = ln x + (x + 1)3             âñå êîðíè
 24    0 = 2, 2x − 2x                  íàèìåíüøèé êîðåíü




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