maths.free › Numerical Methods › Interpolation and Taylor approximation

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.

Töötatud näide · taylor series of e^x

Calculus · Taylor-Series

Taylor series of e^(x)

e^{x}

Samm-sammult

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.

Vastuse avalikustamine

\frac{x^{5}}{120} + \frac{x^{4}}{24} + \frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1 + O((x-0)^{6})

Nüüd sina. Vali probleem, või kirjuta või joonista oma. Iga samm, pilt, vastus peidetud, kuni küsid.

Hoia oma töö

Tasuta konto lisab märkmeid igale õppetunnile, ülestähenduse, oma lahendatud probleemide kohta ühes kohas ja juhendaja kohta, kellelt võid küsida selle lehekülje kohta. Matemaatika ise on avatud kõigile, allkirjastatud või mitte.

Registreeru Sisselogimine

Siin kasutatavad sümbolid

Koputage kõiki sümboleid, mis on mõeldud täieliku definitsiooni, pildi ja iga tähe all.

Kuidas: 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.

Küsimusi, mida inimesed küsivad

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.

Rohkem Numerical Methods

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