- #1
opticaltempest
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I have the the follow 3rd order polynomial approximation for e^3x
[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 believe 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]
[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 believe 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]
Last edited: