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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.

مثال عملي · taylor series of sin(x)

Calculus · Taylor-Series

Taylor series of sin(x)

\sin{\left(x \right)}

خطوة خطوة

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.

كشفت الإجابة

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

الآن أنت اختار مشكلة، أو طبعها أو رسمها بنفسك، كل خطوة، صورة، الإجابة مخفية حتى تسأل.

إبقي عملك الخاص

إن الحساب المجاني يضيفه ملاحظات على كل درس، وسجل لما أنجزته، ومشاكلك التي حلتها في مكان واحد، ومعلم يمكنك أن تسأله عن هذه الصفحة. والرياضيات نفسها مفتوحة أمام الجميع، سواء كانوا مستخدمين أو غير مستخدمين.

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لمس أي رمز للحصول على التعريف الكامل، والصورة، وما يعنيه كل حرف في ذلك.

كيف: 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.

أسئلة يطرحها الناس

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.

أكثر في Real Analysis

Sequences and their limits Convergence of series Improper integrals Functions of a complex variable Continuity and differentiability, rigorously Sequences of functions and uniform convergence