# Help with Taylor's Theorem to obtain error of approximation.

1. Mar 27, 2006

### opticaltempest

I need to use Taylor's Theorem to obtain the upper bound for the error of the approximation on the following

$$e^{\frac{1}{2}} \approx 1 + \frac{1}{2} + \frac{{\left( {\frac{1}{2}} \right)^2 }}{{2!}} + \frac{{\left( {\frac{1}{2}} \right)^3 }}{{3!}}$$

Here is an example problem in the textbook I am following.
http://img157.imageshack.us/img157/8564/100027rh.jpg [Broken]

Here is my work. Am I on the right track?

$$e^{\frac{1}{2}} = 1 + \frac{1}{2} + \frac{{\left( {\frac{1}{2}} \right)^2 }}{{2!}} + \frac{{\left( {\frac{1}{2}} \right)^3 }}{{3!}} + R_3 \left( {\frac{1}{2}} \right)$$

$$e^{\frac{1}{2}} = 1 + \frac{1}{2} + \frac{{\left( {\frac{1}{2}} \right)^2 }}{{2!}} + \frac{{\left( {\frac{1}{2}} \right)^3 }}{{3!}} + \frac{{f^4 (z)}}{{4!}}\left( {\frac{1}{2}} \right)^4$$

where $$0 < z < \frac{1}{2}$$

$$f^4 (z) = e^z$$

$$\frac{{e^0 }}{{4!}}\left( {\frac{1}{2}} \right)^4 < \frac{{e^z }}{{4!}}\left( {\frac{1}{2}} \right)^4 < \frac{{e^{\frac{1}{2}} }}{{4!}}\left( {\frac{1}{2}} \right)^4$$

So the upper bound for the error on the approximation of $$\[ e^{\frac{1}{2}}$$ is

$$\frac{{e^{\frac{1}{2}} }}{{4!}}\left( {\frac{1}{2}} \right)^4 = \frac{{e^{\frac{1}{2}} }}{{384}} \approx 0.004$$

Last edited by a moderator: May 2, 2017