# Homework Help: 4th order Maclaurin Polynomial - Help with error approximation

1. Mar 16, 2006

### opticaltempest

I have the the follow 3rd order polynomial approximation for e^3x

$$f(x) = e^{3x} \approx 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3$$

In an earlier part of the problem, I found

$$f\left( {\frac{1}{3}} \right) = e^{3\left( {\frac{1}{3}} \right)}$$

$$\approx 1 + 3\left( {\frac{1}{3}} \right) + \frac{9}{2}\left( {\frac{1}{3}} \right)^2 + \frac{9}{2}\left( {\frac{1}{3}} \right)^3 = \frac{8}{3} \approx 2.667$$

I need to find the upper bound for the error of the above approximation.

$$f(x) = e^{3x} = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3 + R_3 (x)$$

$$\begin{array}{l} R_3 \left( {\frac{1}{3}} \right) = \frac{{f^{\left( 4 \right)} \left( z \right)}}{{4!}}\left( {\frac{1}{3}} \right)^4 \\ {\rm{where }}0 < z < \frac{1}{3} \\ \end{array}$$

$$f^{\left( 4 \right)} (z) = 27e^{3z}$$

So the bounds on my error

$$\frac{{27e^{3(0)} }}{{4!}} < \frac{{27e^{3z} }}{{4!}} < \frac{{27e^{3\left( {\frac{1}{3}} \right)} }}{{4!}}$$

Which means my error for this function when evaluated at 1/3 is
somewhere between 1.125 and ~3.058. This must be wrong
because f(1/3) is approx. 2.67.
(e - 2.67) is approx 0.5.

I beleive the bounds that I found are saying that my error must be
greater than 1.125 and less than 3.058. Where did I go wrong?

Thanks

EDIT: I found some errors while looking over the problem more carefully.
Please let me know if it looks correct now. Thanks. My corrections are
below.

$$f^{\left( 4 \right)} (z) = 81e^{3z}$$

$$\frac{{81e^{3(0)} }}{{4!}}\left( {\frac{1}{3}} \right)^4 < \frac{{81e^{3(z)} }}{{4!}}\left( {\frac{1}{3}} \right)^4 < \frac{{81e^{3\left( {\frac{1}{3}} \right)} }}{{4!}}\left( {\frac{1}{3}} \right)^4$$

$$\approx 0.04167 < \frac{{81e^{3(z)} }}{{4!}}\left( {\frac{1}{3}} \right)^4 < \approx 0.11326$$

Last edited: Mar 16, 2006
2. Mar 16, 2006

### Hurkyl

Staff Emeritus
This still looks wrong.

3. Mar 16, 2006

### opticaltempest

Oops,

(e - 2.67) is approx. 0.05 not 0.5