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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.
Exemplu lucrat · taylor series of e^x
Pas cu pas
- 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.
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Cum să: 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.
Întrebări pe care oamenii le întreabă
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.
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