dstefan
07-01-2008, 05:58 PM
Chapter 5, Problem 1
Find the cubic Taylor approximation of \sqrt{1+x} around 0.
Solution:
The cubic Taylor approximation of the function f(x) around the point 0 is
f(x) \approx f(0) ~+~ x f'(0) ~+~ \frac{x^2}{2} f''(0) ~+~ \frac{x^3}{6} f^{(3)}(0).
Let f(x) = \sqrt{1+x}. Then f'(x) = \frac{1}{2\sqrt{1+x}}; f''(x) = -\frac{1}{4(1+x)^{3/2}}; f^{(3)}(x) = \frac{3}{8(1+x)^{5/2}} and therefore f(0) = 1; ~f'(0) = \frac{1}{2}; ~f''(0) = -\frac{1}{4}; ~f^{(3)}(0) = \frac{3}{8}
We conclude that
\sqrt{1+x} ~\approx~ 1 ~+~ \frac{x}{2} ~-~ \frac{x^2}{8} ~+~ \frac{x^3}{16}
Find the cubic Taylor approximation of \sqrt{1+x} around 0.
Solution:
The cubic Taylor approximation of the function f(x) around the point 0 is
f(x) \approx f(0) ~+~ x f'(0) ~+~ \frac{x^2}{2} f''(0) ~+~ \frac{x^3}{6} f^{(3)}(0).
Let f(x) = \sqrt{1+x}. Then f'(x) = \frac{1}{2\sqrt{1+x}}; f''(x) = -\frac{1}{4(1+x)^{3/2}}; f^{(3)}(x) = \frac{3}{8(1+x)^{5/2}} and therefore f(0) = 1; ~f'(0) = \frac{1}{2}; ~f''(0) = -\frac{1}{4}; ~f^{(3)}(0) = \frac{3}{8}
We conclude that
\sqrt{1+x} ~\approx~ 1 ~+~ \frac{x}{2} ~-~ \frac{x^2}{8} ~+~ \frac{x^3}{16}