What is the approximate value of 26 up to 3 decimal places using Taylor's theorem?

Answer 1

To approximate ( e^{26} ) using Taylor's theorem, we can expand ( e^x ) around ( x = 0 ) using its Maclaurin series:

[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots ]

For ( x = 26 ), summing the first few terms gives:

[ e^{26} \approx 1 + \frac{26}{1} + \frac{26^2}{2} + \frac{26^3}{6} \approx 1 + 26 + 338 + 4565 \approx 4930 ]

This is a rough estimation, and for more accuracy, additional terms would need to be included or numerical methods employed for precise calculation.