Question

i) Calculate the third-degree Taylor polynomial about 0 x0 = for 2/1
f (x) = 1( + x) .
ii) Use the polynomial in part (i) to approximate 1.1 and find a bound for the
error involved.
iii) Use the polynomial in part (i) to approximate ∫
+
1.0
0
/1 2
1( x) dx .

Answer 1

Answer on Question #46697 – Math – Algorithms | Quantitative Methods

Problem.

i) Calculate the third-degree Taylor polynomial about $0 \times 0 =$ for $2/1$ $f(x) = 1(+x)$.

ii) Use the polynomial in part (i) to approximate 1.1 and find a bound for the error involved.

iii) Use the polynomial in part (i) to approximate $\int$

1.0

/1 2

$1(x) \, dx$

Solution:

The question is incorrectly formatted, so we suppose that we have function $f(x)$ and point $x_0$.

i) The third third-degree Taylor polynomial about $x_0$ equals

$$P(x) = f(x_0) + \frac{f'(x_0)}{1!}(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \frac{f'''(x_0)}{3!}(x - x_0)^3.$$

ii) The approximation of $f(x)$ by the polynomial $P(x)$ in the point $a$ equals

$$P(a) = f(x_0) + \frac{f'(x_0)}{1!}(a - x_0) + \frac{f''(x_0)}{2!}(a - x_0)^2 + \frac{f'''(x_0)}{3!}(a - x_0)^3$$

The error equals $R_3(a) = \frac{f^{(4)}(c)}{4!}(a - x_0)^4$, where $c$ is constant between $x_0$ and $a$. Hence to find to find error we need to find maximum and minimum value of the function $\frac{f^{(4)}(c)}{4!}(a - x_0)^4$ ($c$ is variable) on between $x_0$ and $a$.

iii) The $\int f(x) \, dx$ could be approximate with

$$\begin{aligned} \int_{a}^{b} P(x) \, dx &= \int_{a}^{b} f(x_0) + \frac{f'(x_0)}{1!}(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \frac{f'''(x_0)}{3!}(x - x_0)^3 \, d(x - x_0) \\ &= f(x_0)(x - x_0) + \frac{f'(x_0)}{2!}(x - x_0)^2 + \frac{f''(x_0)}{3!}(x - x_0)^3 + \frac{f'''(x_0)}{4!}(x - x_0)^4 \Bigg|_{a}^{b} \\ &= f(x_0)(b - x_0) + \frac{f'(x_0)}{2!}(b - x_0)^2 + \frac{f''(x_0)}{3!}(b - x_0)^3 + \frac{f'''(x_0)}{4!}(b - x_0)^4 \\ &\quad - \left( f(x_0)(a - x_0) + \frac{f'(x_0)}{2!}(a - x_0)^2 + \frac{f''(x_0)}{3!}(a - x_0)^3 + \frac{f'''(x_0)}{4!}(a - x_0)^4 \right). \end{aligned}$$

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