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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 · taylor series of sin(x)

Calculus · Taylor-Series

Taylor series of sin(x)

\sin{\left(x \right)}

قدم ب قدم

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.

جواب کھوليں

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

اب آپ مسئلہ منتخب کريں يا ٹائپ کريں يا خود کريں ہر قدم ، تصوير ، جواب پوشيدہ هے جب تک آپ سوال نہ کريں

آپ کے اپنے کام رکھیں

مفت اکاؤنٹ تمام درسوں پر نوٹ شامل کر تا هے ، آپ نے جو ختم کر يا هے اس کا ريکورڈ ، آپ کے حل کيے گئے مسئلے ایک جگہ پر اور آپ اس صفحے کے بارے ميں سوال کر سکتے هيں تو تيرچر ريتھ ميں خود تمام لوگوں کے ليے کھلا هے ، خواہ آپ داخل هے يا نہيں

رجسٹر لاگ ان

علامتیں

تمام تعريف کے ليے ، تصوير کے ليے اور اس ميں ہر حروف کا مطلب کے ليے کو ئي علامت کو ٹک کريں

کیسے: 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.

سوالات لوگ پوچھتے ہیں

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.

میں زیادہ 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