Question

Use a Maclaurin series to approximate the integral (in comments) so that the abs(error)<10^-10

Answer 1

\[\int\limits_{0}^{0.1} \cos(x^3) dx \]
so that \[\left| error \right|<10^{-10}\]
You may use the fact that \[\cos x=\sum_{n=0}^{\infty} (-1)^n \frac{ x^{2n} }{ (2n)! }\]

Answer 2

I have no idea how to do this

Answer 3

Do you have a calculus textbook?

Answer 4

Here's a hint - the alternating series remainder theorem tells you the upper bound of the partial sum.

Answer 5

\[\sum_{n=0}^{\infty} \frac{ (-1)^n(0.1)^{6n+1} }{ (6n+1)(2n)! }-\sum_{n=0}^{\infty} 0\]
i get to here. what do i do after that

Answer 6

Simply evaluate the partial sum, if the integral is in fact correct. Because the series is alternating in nature, you know that the value \(S_n\) can be off by no more than \(a_{n+1}\)