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#1 2007-04-25 07:10:57

quackensack
Member
Registered: 2007-02-27
Posts: 47

Maclaurin Series problem

If anyone could explain how the following problem is done, it would be greatly appreciated.

Write the Maclaurin series for cos(x^2).  By antidifferentiating its first three (nonzero) terms, obtain an approximation for the integral of cos(x^2) dx from .6 to 1.  Is this approximation larger than or smaller than the actual value of the integral of cos(x^2) from .6 to 1?  How does the series itself tell you the answer to this question?

I've done the first part which is write out the Maclaurin series for cos(x^2):
cos(x^2) = 1 - x^4/2! +x^8/4! - x^12/6!...
And I've antidifferentiated the first three terms to get:
x - x^5/10 + x^9/216...

However, I'm confused about how to obtain an approximation for the integral of cos(x^2) dx from .6 to 1.

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#2 2007-04-25 07:50:04

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Maclaurin Series problem

That's the easy bit! All you need to do from there (assuming the rest of it is right) is find the value of x - x^5/10 + x^9/216 when x=1, and then take away the value for when x=0.6.

That should get you an answer of ≈ 0.312.

Edit: And to answer the other part, this approximation will be larger than the actual value because the first term in the Maclaurin expansion that you ignore is negative.


Why did the vector cross the road?
It wanted to be normal.

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