Thread: error factor of Taylor Series

the function is:

$\displaystyle \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:

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

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

$\displaystyle \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.

Originally Posted by lllll

the function is:

$\displaystyle \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:

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

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

$\displaystyle \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 ....?

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

How do you know how many factors to expand out?

if you expand it to 5 you get it:

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

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