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[xn] ≤ xn ≤ [xn]+1

(1 + 1/([xn]+1))[xn] < (1 + 1/xn)xn ≤ (1 + 1/[xn])[xn]+1

(1 + 1/([xn]+1))[xn]+1 ⋅ (1/(1+1/([xn+1]))) < (1 + 1/xn)xn < (1 + 1/[xn])[xn] ⋅ (1+1/[xn])

e e

e

y = -x

(1 – 1/y)-y = ( (y-1)/y )-y = ( y/(y-1) )y = ( (y-1+1)/(y-1) )y = (1 + 1/(y-1) )y =

= (1 + 1/(y-1) )y-1⋅(1+1/(y-1)) → e⋅1 = e

z = 1/x

limz→0 (1+z)1/z = e

limx→0 (1+x)1/x = e

limx→0 (ln(1+x))/x = limx→0 (1/x)⋅ln(1+x) = limx→0 ln(1+x)1/x = ln (limx→0(1+x)1/x)= ln e = 1

limx→0 (loga(1+x))/x = limx→0 (ln (1+x))/(x⋅ln a) = 1/ln a

limx→0 (ax - 1)/x = limy→0 y/(loga (1+y)) = ln a

limx→0 (ex - 1)/x = 1

limx→0 (eαln (1+x) - 1)/x = limx→0 (eαln (1+x) - 1)/α⋅ln (1+x) ⋅ (α⋅ln (1+x))/x = α

|sin x| ≤ x

x > 0, B C

0 ≤ sinx ≤ x tg x

0 ≤ (sin x)/x ≤ 1 O A

SOAB = S∆OAC

x/2 ≤ ½ ⋅ 1 ⋅ tg x

x ≤ tg x

x ≤ sinx/cosx,

cosx ≤ sinx/x ≤ 1

1 1 1

Производные

s(t) s(t+h)

s(t), s(t+h), h – приращение.

s’(t) = v(t).


α


a F Q b

(a,b) f(x)

x0 ϵ (a,b)

f(x) = xα, x > 0

x0 > 0

limh→0((x0 + h)α – x0α)/h = limh→0(x0α⋅(1 + h/x0)α – x0α)/h = α ⋅ x0α -1

∃ f’(x0), g’(x0), g(x0) ≠ 0 ==>

∃ (f/g)’(x0) = ( f’(x0)⋅g(x0) – f(x0)⋅g’(x0) )/(g’(x0))2

limh→0 [f(x0+h)/g(x0+h) – f(x0)/g(x0)]⋅1/h =

= limh→0(1/h)⋅[( f(x0+h)⋅g(x0) – f(x0)⋅g(x0+h) )/g(x0+h)⋅g(x0)] =

= limh→0 (1/h)⋅[( f(x0+h)⋅g(x0) – f(x0)⋅g(x0) + f(x0)⋅g(x0) – f(x0)⋅g(x0+h) )/g(x0+h)⋅g(x0)] =

= limh→0 [(f(x0+h) – f(x0))/h ⋅ g(x0) – (g(x0+h) – g(x0))/h ⋅ f(x0)] ⋅ 1/(g(x0+h)⋅g(x0))

k=0n ak⋅xk]’ = Σk=1n ak⋅kxk-1

(tg x)’= (sin x/cos x)’ = =

Дифференцирование

f(a) = b


b

a h

f(a+h) – (α⋅h + β) = ϕ(h) ϕ(h)/h –h→0→0 = f(a) – β, β = f(a) ==>

f(a+h) – f(a) – α⋅h = ϕ(h) (f(a+h) – f(a) – α⋅h)/h = (f(a+h) – f(a))/h – α –h→0→ 0 ==>

α = f’(a), т.е. существует «хорошее» линейное приближение <==> существует f’(a) и это линейное приближение имеет вид f’(a)⋅h + f(a) ≈ f(a+h).

Верно и обратное: пусть существует производная, тогда существует такое приближение.

f’(a)⋅h= α⋅h

ϕ(h) = f(a+h) – f’(a)⋅h – f(a) = [f(a+h) – f(a)] – f’(a)⋅h

ϕ(h)/h = ( f(a+h) – f(a) )/h – f’(a) –h→0→0

f(x), aϵD,

∃ δ > 0 : |x-a| < δ,

x ϵ D ==> f(x) < f(a)

Теорема Ферма

(a,b) f(x)

c ϵ (a,b) – лок. экстремум.

Если в точке с существует производная, то эта производная должна равняться нулю:

∃ f’(c) ==> f’(c) = 0.

Доказательство:

f’(c) = limx→c (f(x)-f(c))/(x-c)

Пусть с – локальный максимум, тогда

справа: f(x) ≤ f(c), x>c,

limx→c+0 (f(x)-f(c))/(x-c) ≤ 0

слева: f(x) ≤ f(c), x<c, ==> f’(c) = 0.

limx→c–0 (f(x)-f(c))/(x-c) ≥ 0

∃ f’(c)

Пусть f(x) определена на [a,b] и

1. f(x) – непрерывна;

2. ∃ f’(x) на (a,b); ==> ∃ ξ ϵ (a,b) : f’(ξ) = 0

3. f(a) = f(b);

f(x) непрерывна на (a,b), с ϵ (a,b),

∀ x ≠ c ∃ f’(x)

(a,c) f’(x) ≥ 0 c – точка max

(c,b) f’(x) ≤ 0

∀ x ≠ c ∃ f’(x)

(a,c) f’(x) ≤ 0 c – точка min

(c,b) f’(x) ≥ 0

f’’(x) = (f’)’(x)

f(n+m)(x) = [f(n)](m)(x)

[α⋅f(x) + β⋅g(x)](n) = α⋅f(n)(x) + β⋅g(n)(x)

Теорема:

Предположим, что f(x), g(x) определена на (a,b), ∃ f(n)(x), g(n)(x). Тогда

(f⋅g)(n)(x) = Σk=0n cnk ⋅ f(k)(x) ⋅ g(n-k)(x).

Доказательство:

Докажем по индукции:

(f⋅g)’(x) = f’(x) ⋅ g(x) + f(x) ⋅ g’(x)

(f⋅g)(n+1)(x) = d/dx (f⋅g)(n)(x) = [(f⋅g)(n)]’(x) = [Σk=0n cnk ⋅ f(k)(x) ⋅ g(n-k)(x)]’ =

= Σk=0n cnk ⋅ [f(k+1)(x) ⋅ g(n-k)(x) + f(k)(x)⋅g(n-k+1)(x)] =

= Σk=0n cnk ⋅ f(k+1)(x) ⋅ g(n-k)(x) + Σk=0n cnk ⋅ f(k)(x) ⋅ g(n-k+1)(x)

m = k+1:

Σk=0n cnk ⋅ f(k+1)(x) ⋅ g(n-k)(x) = Σm=1n+1 cnm-1 ⋅ f(m)(x) ⋅ g(n+1–m)(x)

k = m:

Σm=1n+1 cnm-1 ⋅ f(m)(x) ⋅ g(n+1–m)(x) + Σk=0n cnk ⋅ f(k)(x) ⋅ g(n-k+1)(x) =

= Σk=1n+1 cnk-1 ⋅ f(k)(x) ⋅ g(n+1–k)(x) + Σk=0n cnk ⋅ f(k)(x) ⋅ g(n-k+1)(x) =

= f(x)⋅g(n+1)(x) + f(n+1)(x)⋅g(0)(x) + Σk=n [cnk + cnk-1]⋅ f(k)(x) ⋅ g(n-k+1)(x) =

= Σk=0n cnk ⋅ f(k)(x) ⋅ g(n-k)(x).

f(x), g(x) g(x) ≠ 0 (x ≠ a)

f(x)/g(x) –xa→0 ~ f(x) = o(g(x)) – функция f(x) при x→a стремится к нулю быстрее, чем g(x)

Свойства:

f1(x) = o(g(x))

f2(x) = o(g(x))

1. [f1(x) + f2(x)]/g(x) = f1(x)/g(x) + f2(x)/g(x) → 0

2. f1(x) + f2(x) = o(g(x)) + o(g(x)) = o(g(x))

3. f(x)⋅h(x) = o(g(x)), |h(x)| ≤ M

|f(x)|/|g(x)| ≤ const ~ f(x) = O(g(x)) – порядок малости одинаков у функций f(x) и g(x)

f(x) (a-δ; a+δ), δ>0

∃ f(n)(a).

Формула Тейлора

f(x) = Σk=0n (f(k)(a))/k! (x-a)k = rn(x)

rn(x) = o((x-a)n), limx→a rn(x)/(x-a)n = 0

Доказательство:

Pn(x) = Σk=0n (f(k)(a))/k! ⋅ (x-a)k

Pn(a) = k!⋅(f(k)(a))/k! = f(k)(a)

rn(k)(a) = 0, k ≤ n ==> rn(x) = o((x-a)n):

n=1. r1(x) = f(x) – f(a) – f’(a)⋅(x-a) ==> r1’(a) = 0 => r1(x) = o(x-a)

r1(x)/(x-a) = (f(x)–f(a))/(x-a) – f’(a) –x→a→ 0

n=n-1. rn’(x) = ϕ(x), ϕ(k)(x) = rn(k+1)(x), x=a => rn(k+1)(x) = 0, k+1 ≤ n, k ≤ n-1

ϕ(x)/(x-a)n-1x→a→ 0

rn(x)/(x-a)n = (rn(x) – rn(a))/(x-a)n = rn’(ξ)⋅(x-a)/(x-a)n = rn’(ξ)/(x-a)n-1 =

= ϕ(ξ)/(ξ-a)n-1 ⋅ (ξ-a)n-1/(x-a)n-1, ξ ϵ (a,x), ϕ(ξ)/(ξ-a)n-1 →0, |(ξ-a)n-1/(x-a)n-1| < 1 ==>

==> ϕ(ξ)/(ξ-a)n-1 ⋅ (ξ-a)n-1/(x-a)n-1x→a→ 0 ⋅ 1/ℰ = 0.

Разложение по форме Тейлора с остатком в виде Пеано

f’(a) = f’’(a) = … = f(k-1)(a) = 0

f(k)(a) ≠ 0, k ≤ n

f(x) = f(a) + 0⋅(x-a) + 0⋅(x-a)2 + … + 0⋅(x-a)k-1 = f(k)(a)/k! ⋅ (x-a)k + o((x-a)k).

f(x) – f(a) = (x-a)k ⋅[f(k)(a)/k! + o((x-a)k)/(x-a)k]

k – нечётное ==> нет экстремума

sgn = sgn f(k)(a) k – чётное ==> есть экстремум

f(k)(a) > 0 ==> f(x) – f(a) ≥ 0 – min

f(k)(a) < 0 ==> f(x) – f(a) ≤ 0 – max

f(x)/g(x) = [f(n)(a)/n! ⋅ (x-a)n + o((x-a)n)] / [g(m)(a)/m! ⋅ (x-a)m + o((x-a)m)] =

= [f(n)(a)/n! ⋅ (x-a)n–m + o((x-a)n)/(x-a)m] / [g(m)(a)/m! + o((x-a)m)/(x-a)m]

n=m. f(x)/g(x) = [f(n)(a)/n! + o((x-a)n)/(x-a)n] / [g(m)(a)/m! + o((x-a)m)/(x-a)m]

Вычисление с погрешностью

rn(x) = (x-a)p⋅H(x,a,p)

f(x) = Σk=0n (f(k)(a))/k! (x-a)k + (x-a)p⋅H(x,a,p)

ϕ(t) = Σk=0n (f(k)(t))/k! (x-t)k + (x-t)p⋅H(x,a,p), t ϵ (a–δ, a+δ),

∃ ϕ’(t)

ϕ(a) = f(x) t

ϕ(x) = f(x) |––––––––––––|

∃ ξ ϵ (a,x) : ϕ’(ξ) = 0 a x

ϕ’(t) = Σk=0n (f(k+1)(t))/k! (x-t)k + (-1)⋅k⋅(x-t)k-1⋅ f(k)(t))/k! + (-1)⋅p⋅(x-t)p-1⋅H(x,a,p) =

= f’(t) + f’’(t)⋅(x-t) – f’(t) + f’’’(t)/2! (x-t)2 – f’’(t)/2! ⋅2⋅(x-t) + …

… + f(n+1)(t)/n! (x-t)n – f(n)(t)/n! ⋅n⋅(x-t)n-1 – p⋅(x-t)p-1 ⋅H(x,a,p) =

= f(n+1)(t)/n! (x-t)n – p⋅(x-t)p-1⋅H(x,a,p)

ξ = a + θ⋅(x–a), 0 < θ < 1

f(n+1)(a + θ⋅(x–a))/n! (x–a–θ⋅(x–a))n = p⋅(x–a–θ⋅(x–a))p-1⋅H(x,a,p)

f(n+1)((x–a)(1–θ))/n! ((x–a)(1–θ))n = p⋅((x–a)(1–θ))p-1⋅H(x,a,p)

H = [f(n+1)((x–a)(1–θ)) / (n!⋅p)] ⋅ ((x–a)(1–θ))n+1–p

(x-a)p⋅H = [f(n+1)((x–a)(1–θ)) / (n!⋅p)] ⋅(1–θ)n+1–p (x-a)n+1 < погрешности, зависящей от n.

остаток формулы Тейлора в форме Лагранжа: f(n+1)(a + θ⋅(x–a))/(n+1)! ⋅(x-a)n+1

остаток формулы Тейлора в форме Коши: f(n+1)(a + θ⋅(x–a))/n! ⋅(1–θ)n⋅(x-a)n

= (1+29)1/3= = 3 = 3⋅(1+1/9)1/3 = 3⋅[1 + 1/3⋅x – 1/3⋅2/3⋅1/2⋅1⋅x2 + 1/3⋅2/3⋅5/3⋅1/(1+θx)8/3⋅1/3!⋅x3] = 3⋅[ 1 + 1/27 – 1/729 + 5/95⋅1/(1+θx)8/3] ;

5⋅3/95 = 5/3 ⋅ 1/94 < 1/3000;

≈ 3 + 1/9 – 1/243 ≈ 3.10699

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