error factor of Taylor Series

**lllll** · April 11th 2008, 11:25 PM

the function is:

$\int^{1}_{0} x cos(x^3) \ dx$ find the error to 3 decimal places.

so I expressed cos as a Taylor Series which gives me:

$\int^{1}_{0} x \sum^{\infty}_{n=0} -1^n \frac{x^{6n}}{(2n)!} \ dx$

$\int^{1}_{0} \sum^{\infty}_{n=0} -1^n \frac{x^{6n+1}}{(2n)!} \ dx$

$\sum^{\infty}_{n=0} -1^n \frac{x^{6n+2}}{(6n+2)(2n)!} \bigg{|}^{1}_{0} = \sum^{\infty}_{n=0} \frac{-1^n}{(6n+2)(2n)!}$

but to find the error to whatever decimal place I get stuck, if someone could show me how to find it, and explain the process behind their approach that would really help me out.

**mr fantastic** · April 12th 2008, 12:08 AM

Originally Posted by lllll

the function is:

$\int^{1}_{0} x cos(x^3) \ dx$ find the error to 3 decimal places.

so I expressed cos as a Taylor Series which gives me:

$\int^{1}_{0} x \sum^{\infty}_{n=0} -1^n \frac{x^{6n}}{(2n)!} \ dx$

$\int^{1}_{0} \sum^{\infty}_{n=0} -1^n \frac{x^{6n+1}}{(2n)!} \ dx$

$\sum^{\infty}_{n=0} -1^n \frac{x^{6n+2}}{(6n+2)(2n)!} \bigg{|}^{1}_{0} = \sum^{\infty}_{n=0} \frac{-1^n}{(6n+2)(2n)!}$

but to find the error to whatever decimal place I get stuck, if someone could show me how to find it, and explain the process behind their approach that would really help me out.

Reading this thread: http://www.mathhelpforum.com/math-he...ion-error.html

might help ....?

**lllll** · April 12th 2008, 10:02 AM

$\sum^{\infty}_{n=0} \frac{-1^n}{(6n+2)(2n)!} <0.0001$

How do you know how many factors to expand out?

if you expand it to 5 you get it:

$\frac{1}{2}-\frac{1}{16}+\frac{1}{336}-\frac{1}{14400}+\frac{1}{1048320} < 0.0001$

so taking the first 3 terms I get $\frac{37}{84} = 0.4404762$ and since I rejected the 4th term will my error be $\leq \left| \frac{-1^3}{(6(3)+2)(2(3))!} \right| = |- 0.000069444| = 0.000069444$

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