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Taylor approximation and error

How well a polynomial stands in for a function, and how far you can trust it.

Near the centre the Taylor polynomial is nearly the function; the error is controlled by the next derivative and the distance from the centre. This is why calculators can compute sin: a handful of terms inside a small interval, plus symmetry.

例題 · taylor series of sin(x)

Calculus · Taylor-Series

Taylor series of sin(x)

\sin{\left(x \right)}

ステップバイステップ

  1. f(x) = \sin{\left(x \right)}

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

  2. f^{(0)}(0) = \sin{\left(x \right)}\big|_{x=0} = 0

    Derivative 0 at the centre.

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

    Derivative 1 at the centre.

  4. f^{(2)}(0) = - \sin{\left(x \right)}\big|_{x=0} = 0

    Derivative 2 at the centre.

  5. f^{(3)}(0) = - \cos{\left(x \right)}\big|_{x=0} = -1

    Derivative 3 at the centre.

  6. f^{(4)}(0) = \sin{\left(x \right)}\big|_{x=0} = 0

    Derivative 4 at the centre.

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

    Derivative 5 at the centre.

  8. \frac{x^{5}}{120} - \frac{x^{3}}{6} + x

    Assemble the terms up to degree 5.

答えを表示
\frac{x^{5}}{120} - \frac{x^{3}}{6} + x + O((x-0)^{6})

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どうやって: Taylor approximation and error

  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.

よくある質問

What is the ε, δ definition actually saying?

That you can make the output as close to the limit as anyone demands (within ε) by keeping the input close enough (within δ). It replaces "approaches" with a challenge-and-response that can be checked.

Why does the harmonic series diverge when its terms go to zero?

Because the terms shrink too slowly: group them as 1/3 + 1/4 > 1/2, 1/5 + … + 1/8 > 1/2, and so on: infinitely many halves.

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