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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.

Radni primjer · taylor series of sin(x)

Calculus · Taylor-Series

Taylor series of sin(x)

\sin{\left(x \right)}

Korak po korak

  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.

Otkrij odgovor
\frac{x^{5}}{120} - \frac{x^{3}}{6} + x + O((x-0)^{6})

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Simboli korišteni ovdje

Dodirnite bilo koji simbol za potpunu definiciju, sliku i što svako slovo u njemu znači.

Kako: 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.

Pitanja koja ljudi postavljaju

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.

Više u Real Analysis

Sequences and their limitsConvergence of seriesImproper integralsFunctions of a complex variableContinuity and differentiability, rigorouslySequences of functions and uniform convergence