# Another Taylor series

• Nov 10th 2008, 07:16 PM
fastcarslaugh
Another Taylor series
Let https://webwork.math.uga.edu/webwork...2faa4a6411.png.
Find the MacLaurin polynomial of degree 5 for https://webwork.math.uga.edu/webwork...20658adfd1.png.

i know that this can be solved by using this as a base

then I would just plug in (-2t^4) in place of x and expand it to 5th degree.

then I think I messed up here.

• Nov 11th 2008, 12:18 AM
CaptainBlack
Quote:

Originally Posted by fastcarslaugh
Let https://webwork.math.uga.edu/webwork...2faa4a6411.png.
Find the MacLaurin polynomial of degree 5 for https://webwork.math.uga.edu/webwork...20658adfd1.png.

i know that this can be solved by using this as a base

then I would just plug in (-2t^4) in place of x and expand it to 5th degree.

then I think I messed up here.

Not quite, you need to observe that:

$F'(x)=e^{-2x^4}$

Then use the series expansion of $e^x$ to expand this , then integrate term by term to get the desired series.

CB
• Nov 11th 2008, 12:36 PM
fastcarslaugh
I seem to be integrating this incorrectly..
• Nov 11th 2008, 06:13 PM
skeeter
$e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + ...$

$e^{-2t^4} \approx 1 - 2t^4$

$F(x) \approx \int_0^x 1 - 2t^4 \, dt = \left[t - \frac{2}{5}t^5\right]_0^x = x - \frac{2}{5}x^5$