Approximation of \( \sqrt[4]{17} \) using Taylor Expansion:
Choose \( x_0 = 16 \) because \( \sqrt[4]{16} = 2 \). Write:
$$17 = 16 + 1 = x_0 + \Delta x, \quad \text{where } x_0 = 16, \ \Delta x = 1.$$
The function is \( f(x) = x^{1/4} \). The 3-term Taylor expansion:
$$f(x) \approx f(x_0) + \Delta x \, f'(x_0) + \frac{\Delta x^2}{2} f''(x_0).$$
Table of derivatives:
| n |
\( f^{(n)}(x) \) |
\( f^{(n)}(x_0) \) |
| 0 |
\( x^{1/4} \) |
\( 16^{1/4} = 2 \) |
| 1 |
\( \frac{1}{4} x^{-3/4} \) |
\( \frac{1}{4} \cdot 16^{-3/4} = \frac{1}{32} \) |
| 2 |
\( -\frac{3}{16} x^{-7/4} \) |
\( -\frac{3}{16} \cdot 16^{-7/4} = -\frac{3}{2048} \) |
Insert values into expansion:
$$f(x) \approx 2 + 1 \cdot \frac{1}{32} + \frac{1^2}{2} \cdot \left(-\frac{3}{2048}\right)$$
$$= 2 + 0.03125 - 0.00073242 = 2.03051758.$$
Exact value using calculator: \( \sqrt[4]{17} \approx 2.030543185 \).
Truncation error:
$$E = \lvert \sqrt[4]{17} - S_3(16) \rvert = \lvert 2.030543185 - 2.03051758 \rvert = 2.5607 \times 10^{-5}.$$
Relative error:
$$\frac{E}{\sqrt[4]{17}} = \frac{2.5607 \times 10^{-5}}{2.030543185} \approx 1.261 \times 10^{-5}.$$
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