Approximation of \( \sqrt[3]{26} \) using Taylor Expansion:
Choose \( x_0 = 27 \) because \( \sqrt[3]{27} = 3 \). Write:
$$x = x_0 + \Delta x, \quad x_0 = 27, \ \Delta x = -1.$$
The function is \( f(x) = x^{1/3} \). 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 |
\( x^{1/3} \) |
\( 3 \) |
| 1 |
\( \frac{1}{3} x^{-2/3} \) |
\( \frac{1}{27} \) |
| 2 |
\( -\frac{2}{9} x^{-5/3} \) |
\( -\frac{2}{2187} \) |
Insert values into expansion:
$$f(x) \approx 3 + (-1)\left(\frac{1}{27}\right) + \frac{(-1)^2}{2}\left(-\frac{2}{2187}\right)$$
$$= 3 - 0.037037 - 0.00045724737 = 2.962506.$$
Exact value using calculator: \( \sqrt[3]{26} \approx 2.962496 \).
Truncation error:
$$E = \lvert \sqrt[3]{26} - S_3(27) \rvert = \lvert 2.962496 - 2.962506 \rvert = 9.647185 \times 10^{-6}.$$
Relative error:
$$\frac{E}{\sqrt[3]{26}} = \frac{9.647185 \times 10^{-6}}{2.962496} \approx 3.256438 \times 10^{-6}.$$
For more details, please contact me here.