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Interpolation and Taylor approximation

Fitting polynomials through points, and approximating functions near a centre.

Through n + 1 points passes exactly one polynomial of degree n (Lagrange). Near a point, the Taylor polynomial matches the derivatives. Picture it: the dashed polynomial hugging the curve near the centre and peeling away. Think it: approximation theory asks for the best polynomial over a whole interval, Chebyshev, not Taylor.

Вработен пример · taylor series of e^x

Calculus · Taylor-Series

Taylor series of e^(x)

e^{x}

Чекор по чекор

f(x) = e^{x}
Taylor series about x = 0: f(x) = Σ fᵏ(x₀)/k! · (x − x₀)ᵏ.
f^{(0)}(0) = e^{x}\big|_{x=0} = 1
Derivative 0 at the centre.
f^{(1)}(0) = e^{x}\big|_{x=0} = 1
Derivative 1 at the centre.
f^{(2)}(0) = e^{x}\big|_{x=0} = 1
Derivative 2 at the centre.
f^{(3)}(0) = e^{x}\big|_{x=0} = 1
Derivative 3 at the centre.
f^{(4)}(0) = e^{x}\big|_{x=0} = 1
Derivative 4 at the centre.
f^{(5)}(0) = e^{x}\big|_{x=0} = 1
Derivative 5 at the centre.
\frac{x^{5}}{120} + \frac{x^{4}}{24} + \frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1
Assemble the terms up to degree 5.

Одкриј го одговорот

\frac{x^{5}}{120} + \frac{x^{4}}{24} + \frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1 + 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?: Interpolation and Taylor approximation

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

Why not just solve exactly?

Most equations have no closed-form solution at all, and many that do are unusable in practice. A numerical method delivers as many correct digits as you need, and a good one tells you how many that is.

Why can Newton's method fail?

If it starts where the tangent is nearly flat it shoots far away; near a repeated root it slows to a crawl; and with several roots it may land on the wrong one. A bracketing method like bisection is slower but cannot fail.

Повеќе во Numerical Methods

Root finding: bisection and Newton's method Numerical integration: trapezoid and Simpson Floating point and error