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Taylor approximation and error

How well a polynomial stands in for a function, and how far you can trust it.

Near the centre the Taylor polynomial is nearly the function; the error is controlled by the next derivative and the distance from the centre. This is why calculators can compute sin: a handful of terms inside a small interval, plus symmetry.

Вработен пример · taylor series of sin(x)

Calculus · Taylor-Series

Taylor series of sin(x)

\sin{\left(x \right)}

Чекор по чекор

f(x) = \sin{\left(x \right)}
Taylor series about x = 0: f(x) = Σ fᵏ(x₀)/k! · (x − x₀)ᵏ.
f^{(0)}(0) = \sin{\left(x \right)}\big|_{x=0} = 0
Derivative 0 at the centre.
f^{(1)}(0) = \cos{\left(x \right)}\big|_{x=0} = 1
Derivative 1 at the centre.
f^{(2)}(0) = - \sin{\left(x \right)}\big|_{x=0} = 0
Derivative 2 at the centre.
f^{(3)}(0) = - \cos{\left(x \right)}\big|_{x=0} = -1
Derivative 3 at the centre.
f^{(4)}(0) = \sin{\left(x \right)}\big|_{x=0} = 0
Derivative 4 at the centre.
f^{(5)}(0) = \cos{\left(x \right)}\big|_{x=0} = 1
Derivative 5 at the centre.
\frac{x^{5}}{120} - \frac{x^{3}}{6} + x
Assemble the terms up to degree 5.

Одкриј го одговорот

\frac{x^{5}}{120} - \frac{x^{3}}{6} + x + O((x-0)^{6})

Сега ти. Секој чекор, слика, одговорот е скриен додека не прашаш.

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Потпиши се Најава

Симболи кои се користат тука

Допрете било кој симбол за целосната дефиниција, слика, и што значи секоја буква во неа.

Како да го направиш тоа?: Taylor approximation and error

Taylor series about x = 0: f(x) = Σ fᵏ(x₀)/k! · (x − x₀)ᵏ.
Derivative 0 at the centre.
Derivative 1 at the centre.
Derivative 2 at the centre.
Derivative 3 at the centre.
Derivative 4 at the centre.
Derivative 5 at the centre.
Assemble the terms up to degree 5.

Прашања што луѓето ги поставуваат

What is the ε, δ definition actually saying?

That you can make the output as close to the limit as anyone demands (within ε) by keeping the input close enough (within δ). It replaces "approaches" with a challenge-and-response that can be checked.

Why does the harmonic series diverge when its terms go to zero?

Because the terms shrink too slowly: group them as 1/3 + 1/4 > 1/2, 1/5 + … + 1/8 > 1/2, and so on: infinitely many halves.

Повеќе во Real Analysis

Sequences and their limits Convergence of series Improper integrals Functions of a complex variable Continuity and differentiability, rigorously Sequences of functions and uniform convergence