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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.
Toimiva esimerkki · taylor series of sin(x)
Askel kerrallaan
- 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.
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Miten: 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.
Kysymyksiä
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.
Lisää Real Analysis
Sequences and their limitsConvergence of seriesImproper integralsFunctions of a complex variableContinuity and differentiability, rigorouslySequences of functions and uniform convergence