Find a Taylor polynomial and estimate the error

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Math Workspace (CAS)

CAS keeps the derivative computations, coefficient extraction, polynomial evaluation, and error check in a single readable transcript. Each step can be verified or extended directly in the workspace.

Why Taylor polynomials matter

Most functions are not polynomials, but polynomials are easy to evaluate. Taylor's theorem says that near a chosen centre $a$ , a smooth function can be approximated by a polynomial whose coefficients come from the derivatives of the function at $a$ .

The formula is $T_{n} (x) = \sum_{k = 0}^{n} \frac{f ^{(k)} ( a )}{k !} (x - a)^{k}$ . The approximation gets better as $n$ increases, and Taylor's theorem gives an explicit bound on the error in terms of the next derivative.

Step-by-step: building $T_{5} (x)$ for $sin (x)$ at $a = 0$

The goal is to find the coefficients $a_{k} = f^{(k)} (0) / k!$ for $k = 0, 1, 2, 3, 4, 5$ .

1Compute the derivatives: $sin (x) \to cos (x) \to - sin (x) \to - cos (x) \to sin (x) \to cos (x)$ . The pattern repeats with period 4.
2Evaluate each at $x = 0$ : $0, 1, 0, - 1, 0, 1$ . Even-order derivatives are zero because $sin (0) = 0$ .
3Divide by the corresponding factorials: $a_{0} = 0$ , $a_{1} = 1$ , $a_{2} = 0$ , $a_{3} = - 1/6$ , $a_{4} = 0$ , $a_{5} = 1/120$ .
4Assemble the polynomial: $T_{5} (x) = x - x^{3} /6 + x^{5} /120$ .
5Evaluate at $x = 0.5$ : $T_{5} (0.5) \approx 0.479427$ .
6Compare with $sin (0.5) \approx 0.479426$ . The error is about $1.54 \times 1 0^{- 6}$ .

Checking the error bound

Taylor's theorem says the remainder after $n$ terms satisfies $∣ R_{n} (x) ∣ \leq \frac{M}{( n + 1 )!} ∣ x - a ∣^{n + 1}$ , where $M$ is a bound on $∣ f^{(n + 1)} ∣$ between $a$ and $x$ .

For $sin (x)$ , every derivative has absolute value at most $1$ , so $M = 1$ . After 5 terms the next non-zero term is $x^{7} /5040$ . At $x = 0.5$ , the bound is $(0.5)^{7} /5040 \approx 1.55 \times 1 0^{- 6}$ .

The actual error $1.54 \times 1 0^{- 6}$ is just below this bound, confirming Taylor's theorem and showing that the bound is tight for $sin (x)$ near the origin.

What changes when you pick a different centre or degree

Centring at $a = π /4$ instead of $0$ would produce a polynomial that is accurate near $x = π /4$ but less accurate far from it. The coefficients would involve $sin (π /4)$ and $cos (π /4)$ instead of $0$ and $1$ .

Increasing the degree shrinks the error. Adding the $x^{7} /5040$ term reduces the error from $1 0^{- 6}$ to roughly $1 0^{- 9}$ at $x = 0.5$ . Each extra term buys roughly three more decimal places of accuracy near the origin.

Common Pitfall

The Taylor polynomial is not the same as the Taylor series. The polynomial stops at degree $n$ ; the series is an infinite sum and converges to the function only where the remainder goes to zero.

Try a Variation

Build $T_{3} (x) = x - x^{3} /6$ for $sin (x)$ and estimate the error at $x = 1.0$ . Does the bound still hold? How does reducing the degree affect accuracy further from the centre?

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