Question

{Use of Tech} Best center point Suppose you wish to approximate cos (π/ 2) using Taylor polynomials. Is the approximation more accurate if you use Taylor polynomials centered at 0 or at π/6? Use a calculator for numerical experiments and check for consistency with Theorem 11.2. Does the answer depend on the order of the polynomial?

Accepted Answer

Recall that the Taylor polynomial of a function $$f(x)$$ centered at a point $$a is $$given by the formula:  
$$f(x) \approx \sum_{n=0}^N \frac{f^{(n)}(a)}{n!} (x - a)^n,$$  
where $$f^{(n)}(a$$)$$ is the n$th derivative of f$ evaluated at a$, and N$ is the order of the polynomial. Identify the function and points involved: here, f(x$$) = \cos x$$, and we want to approximate $$\cos(\pi/2)$$. The two centers to compare are a=0$ and a$$=\pi/6$$. Understand that the accuracy of the Taylor polynomial depends on how close the point of approximation x$$=\pi/2$$ is to the center a$. The error term (remainder) in Taylor's theorem involves (x$$ - $$a)^{N+1}$$, so smaller $$|x - a|$$ generally leads to better accuracy for the same order $$N. $$Perform numerical experiments using a calculator or software: compute Taylor polynomials of various orders $$N$$ centered at $$0$$ and at $$\pi/6$$, then evaluate them at $$x=\pi/2. $$Compare the absolute errors $$|\cos(\pi/2) - P_N(\pi/2)|$$ for each center and order. Analyze whether the difference in accuracy depends on the order $$N. $$According to Theorem 11.2 (Taylor's theorem with remainder), the error decreases as $$N$$ increases, but the center closer to $$\pi/2 ($$which is $$\pi/6) $$should generally give a more accurate approximation for lower orders. For very high orders, both centers may yield similar accuracy.

Risposta video verificata per un problema simile:

Concetti chiave

Taylor Polynomials and Centering Point

Remainder Term and Error Bound (Theorem 11.2)

Effect of Polynomial Order on Approximation Accuracy