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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.
उदाहरण · taylor series of e^x
चरणद्वारा चरण
- 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.
जवाफ प्रकट गर्नुहोस्
अब तपाईँ समस्या रोज्नुहोस्, वा टाइप गर्नुहोस् वा तपाईँको आफ्नै रेखाचित्र गर्नुहोस् । प्रत्येक चरण, एउटा तस्वीर, तपाईँले सोध्नु अघि उत्तर लुकेको छ ।
एक निःशुल्क खाता प्रत्येक पाठ मा नोट थप्छ, तपाईं के समाप्त गरेको छ को एक रेकर्ड, एक ठाउँमा आफ्नो समाधान समस्या, र एक शिक्षक तपाईं यो पृष्ठ बारेमा सोध्न सक्नुहुन्छ. गणित आफैलाई सबैलाई खुला छ, मा साइन इन वा छैन.
साइन अप गर्नुहोस् लगइनयहाँ प्रयोग गरिएको प्रतीक
पूर्ण परिभाषा, एउटा तस्वीर, र यो प्रत्येक अक्षर अर्थ के लागि कुनै पनि प्रतीक ट्याप गर्नुहोस्।
कसरी: 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.
प्रश्नहरू मानिसहरूले सोध्छन्
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.
यसमा थप Numerical Methods
Root finding: bisection and Newton's methodNumerical integration: trapezoid and SimpsonFloating point and error