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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.

工作范例 · taylor series of sin(x)

Calculus · Taylor-Series

Taylor series of sin(x)

\sin{\left(x \right)}

一步

f(x) = \sin{\left(x \right)}
Taylor series about x = 0: f(x) = Σ fᵏ(x₀)/k! · (x − x₀)ᵏ.
f^{(0)}(0) = \sin{\left(x \right)}\big|_{x=0} = 0
Derivative 0 at the centre.
f^{(1)}(0) = \cos{\left(x \right)}\big|_{x=0} = 1
Derivative 1 at the centre.
f^{(2)}(0) = - \sin{\left(x \right)}\big|_{x=0} = 0
Derivative 2 at the centre.
f^{(3)}(0) = - \cos{\left(x \right)}\big|_{x=0} = -1
Derivative 3 at the centre.
f^{(4)}(0) = \sin{\left(x \right)}\big|_{x=0} = 0
Derivative 4 at the centre.
f^{(5)}(0) = \cos{\left(x \right)}\big|_{x=0} = 1
Derivative 5 at the centre.
\frac{x^{5}}{120} - \frac{x^{3}}{6} + x
Assemble the terms up to degree 5.

发送答案

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

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如何: Taylor approximation and error

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.

人们对人们提出的问题

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.

Taylor series of sin(x)

一步

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如何: Taylor approximation and error

人们对人们提出的问题

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