I have the the follow 3rd order polynomial approximation for e^3x(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

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

[/tex]

In an earlier part of the problem, I found

[tex]

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

[/tex]

[tex]

\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

[/tex]

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

[tex]

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

[/tex]

[tex]

\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}

[/tex]

[tex]

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

[/tex]

So the bounds on my error

[tex]

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

[/tex]

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.

[tex]

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

[/tex]

[tex]

\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

[/tex]

[tex]

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

[/tex]

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

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