page 96

96 / 102 Russian, pre-1918 confidence 0.88 not checked by a human

Provenance

transcript transcript: 2026-09-29T23:40:07+00:00 machine-generated
normalization normalization: 2026-10-02T18:15:40+00:00 machine-generated

Full TPME chain

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Text as printed modern spelling both

as printed − 4 + modern spelling − 1 +

0 93 0.75
1 71. Вычислимъ тецерь 0.96
2 f(x) =x ³ –15x +40, 0.86
3 2,2+0,1 0.71
4 Прежде всего опредѣлимъ, нѣтъ ли тахітит’а или тіні¬ 0.90
5 тит’а въ предѣлахъ отъ х= 2,2 до х= 2,3. 0.93
6 f’(x) = 3x ²-15 0.81
7 f"(x) =6x 0.88
8 Ирих=–1V5 ’(x) =0. Но 5 заключается внутриданныхъ 0.82
9 пределовъ х, потому что 5 = 2,2360+0,00017. Слѣд. внутри дан¬ 0.89
10 ныхъ предѣловъ х-а f(х) имѣетъ тіпітит при х=5; поэтому 0.90
11 f(V5) будетъ приближеннымъ значеніемъ f(2,2+01ẞ) съ недо¬ 0.95
12 статкомъ, и предѣломъ неизбѣжной погрѣшности будетъ боль- 0.99
13 шая изъ двухъ разностей: (2,2) — (V5) или (2,3)— (V5). 0.90
14 Каждую изъ этихъ разностей можно вычислить или непосред¬ 0.99
15 ственно, или посредствомъ Тайлоровой строчки. 0.99
16 Непосредственное вычисленіе даетъ: 0.99
17 f(2,2) =17,648 0.99
18 f(2,3) = 17,667 0.97
19 f(V5) =17,639+0,001 0.95
20 f(2,2) — f(V5) = 0,009 0.92
21 f(2,3) — f(V5) = 0,028 0.95
22 Итакъ f(2,2+0,13) = f(5) +0,028λ ли 0.84
23 ƒ(2,2+0,1β) =17,639-+-0,02 0.82
24 ƒ(2,2+0,1β) =17,63+-0,04ρ 0.87
25 f(2,2 +0, 13) =17, 6-+0,1σ. 0.85
26 Разсмотримъ опредѣленіе предѣла непзбежной погрешно¬ 0.96
27 сти посредствомъ Тайлоровой строчки. 1.00
28 Hзовемъ 2,3 –V5 =р, V5 – 2,2=; тогда 0.82
29 f (2,3) —f(V5) =pf(V5) + 0.89
30 f(2,2) –f(V5) =qf(V 5) + 0.91
31 1 0.60
32 f'(V5+0) 0.77
33 f’’(V 5 –) 0.71
34 Такъ какъ f’(V5) =0, а f’(x) =6х, то 0.88
35 f(2,3) – f (V 5) = 3 p 2(V 5 +θp) 0.83
36 f(2, 2) –f(V 5) = 3q2 (V 5 –Cq) 0.82
37 Найбольшее значеніе первой разности будетъ при =1, 0.97
38 а второ̆ пр =. оэтому 0.80
39 f(2,3) – f(V 5) = 3p 22, 3 = 3.2,3(2,3 –V 5) "— 0,03 0.83
40 f(2,2) –f(V5) = 3q 2V 5 =3.V 5 (V 5 — 2,2) 0.86