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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.

Megdolgozott példa: taylor series of e^x

Calculus · Taylor-Series

Taylor series of e^(x)

e^{x}

Lépésről lépésre

  1. f(x) = e^{x}

    Taylor series about x = 0: f(x) = Σ fᵏ(x₀)/k! · (x − x₀)ᵏ.

  2. f^{(0)}(0) = e^{x}\big|_{x=0} = 1

    Derivative 0 at the centre.

  3. f^{(1)}(0) = e^{x}\big|_{x=0} = 1

    Derivative 1 at the centre.

  4. f^{(2)}(0) = e^{x}\big|_{x=0} = 1

    Derivative 2 at the centre.

  5. f^{(3)}(0) = e^{x}\big|_{x=0} = 1

    Derivative 3 at the centre.

  6. f^{(4)}(0) = e^{x}\big|_{x=0} = 1

    Derivative 4 at the centre.

  7. f^{(5)}(0) = e^{x}\big|_{x=0} = 1

    Derivative 5 at the centre.

  8. \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.

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

Symbols used here

e
Euler's number
2.71828…, the base whose exponential is its own derivative.
O(n^2),\ \Theta,\ \Omega
big-O notation
Grows no faster than n² (up to a constant), for large n.
\approx
approximately equal
Equal to the precision shown, not exactly.
f'(x),\ \frac{dy}{dx}
derivative
Instantaneous rate of change; slope of the graph.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
x_{n+1} = x_n - \frac{f(x_n)}{f\'(x_n)}
Newton iteration
The next approximation follows the tangent to the axis.

How to: Interpolation and Taylor approximation

  1. Taylor series about x = 0: f(x) = Σ fᵏ(x₀)/k! · (x − x₀)ᵏ.
  2. Derivative 0 at the centre.
  3. Derivative 1 at the centre.
  4. Derivative 2 at the centre.
  5. Derivative 3 at the centre.
  6. Derivative 4 at the centre.
  7. Derivative 5 at the centre.
  8. Assemble the terms up to degree 5.

Questions people ask

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.

Próbáld a sajátodat.

Még több Numerical Methods

Root finding: bisection and Newton's methodNumerical integration: trapezoid and SimpsonFloating point and error