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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.

कार्यरत उदाहरण · taylor series of e^x

Calculus · Taylor-Series

Taylor series of e^(x)

e^{x}

चरणानुक्रमे

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.

उत्तर उघडा

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

आता तुम्ही प्रश्न निवडा, किंवा स्वतःचा प्रश्न टाइप करा किंवा रेखाटून दाखवा. प्रत्येक टप्प्यात, चित्रे आहेत, तुम्ही विचारण्याआधी उत्तर लपवलेले आहे.

स्वतःचे कार्य चालू ठेवा

मोफत खाते प्रत्येक शिकवणी वर टिपण्णी जोडते, तुम्ही पूर्ण केलेल्या गोष्टीचा नोंदवही, तुमच्या सोडविलेल्या समस्या एका ठिकाणी, आणि या पानाच्या विषयी तुम्ही विचारू शकता अशा शिक्षकाला. गणित स्वतःच सर्वांसाठी खुले आहे, दाखलन झालेले असो किंवा नसो.

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कसे कराल: 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.

लोक विचारतात ते प्रश्न

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.

अधिक माहिती Numerical Methods

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