Question

Use the remainder term to find the minimum order of the Taylor polynomial, centered at 0, that is required to approximate the following quantity with an absolute error no greater than 10^-2.

√1.06.
n>= __________

Answer 1

Answer:

n ≥  3

Step-by-step explanation:

Applying the remainder term in evaluating the minimum order of the Taylor polynomial

absolute error ≤ 10^-2

$\sqrt{1.06}$

∴ n ≥   ?

The remainder term is the leftover term after computation ( dividing one polynomial with another )

attached below is the detailed solution

Answer 2

The minimum order of the Taylor polynomial, n≥3

What is Taylor polynomial?

Taylor polynomial is a series of functions that has an infinite sum of terms that are expressed in terms of the function's derivatives.

$\rm f(a)+\frac{f'(a)}{1!} (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +....,$

Applying the Taylor series polynomial, the minimum order of the Taylor polynomial, centered at 0

$\rm f(0)+\frac{f'(0)}{1!} (x-0)+\frac{f''(0)}{2!} (x-0)^{2} +....,$

f(x) = $\sqrt{x+1}$

$f'(x)=\frac{1}{2}$

$f''(x)=3/8$

substituting in the Taylors series

T(x) = $1+\frac{x}{2} -\frac{x^{2} }{8} +\frac{x^{3} }{16}$......

T(0.06) = $1+\frac{0.06}{2} -\frac{0.06^{2} }{8} +\frac{0.06^{3} }{16}...$

T(0.06) =1.03

f(0.06) =

$\sqrt{0.06+1}\\= 1.03$

Therefore, the minimum order of the Taylor polynomial, n≥3

Learn more about Taylor polynomial;

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