Error bound for e^(1/2) using Taylor's theorem with remainder

[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 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 $$\[<br /> e^{\frac{1}{2}}$$ is

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

 

[mfb]

That works. Estimating the error numerically needs a value for e^1/2 which you want to evaluate, of course, but luckily it's just a small correction.

For comparison: The actual error is 0.0029.