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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.

ምሳሌ · taylor series of e^x

Calculus · Taylor-Series

Taylor series of e^(x)

e^{x}

እርምጃ በደረጃ

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.

መልሱን አሳይ

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

አሁን ችግርን ይምረጡ፣ ወይም ጻፉ ወይም በራሳችሁ ቅርጽ አቅርቡ። እርምጃው ሁሉ፣ ምስል፣ መልሱ እስከምትጠይቅ ድረስ የተደበቀ ነው።

የራስዎን ስራዎች ይያዙ

ነጻ የሆኑት መተግበሪያዎች በየክፍል ውስጥ ማስታወሻዎችን ያካትታሉ፣ የተጠናቀቁትን ነገሮች መዝገብ፣ የተፈቱ ችግሮችን በአንድ ቦታ፣ ስለዚህ ገጽ መጠየቅ የሚችሉትን ረዳት ያካትታሉ፡፡ የቁጥር ትምህርት ለሁሉም ሰው ክፍት ነው፣ ተቀባይነት ያለው ወይም የለም፤

ምዝገባ መግባቱ

ፊደላት

ለሙሉ መግለጫ፣ ምስል፣ እና በውስጡ ያለውን ፊደል ትርጓሜ ለማየት ማንኛውንም ምልክት ይጫኑ።

እንዴት: 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.

ጥያቄዎች

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.

በ Numerical Methods

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