1
$\begingroup$

Let $T_k(x)$ be the degree $k$ Taylor polynomial of the function $f(x)=\sin(x)$ at $a=0$. Suppose you approximate $f(x)$ by $T_k(x)$. If $|x|\le 1$, how many terms are needed (that is, what is $k$) to obtain an error less than $\frac 1 {5040}$?

I don't really understand what they are asking or how to get the answer. I was thinking about using alternating series approximation. I am sure how to go about doing that, if that is true.

How do I do this?

Thanks!

$\endgroup$
1
$\begingroup$

When $|x|\le 1$, the Maclaurin series for $\sin x$ is indeed an alternating series. The absolute value of the error when we truncate just after the term in $x^{2n+1}$ has absolute value $\le$ the absolute value of the first "neglected" term. And it is $\lt$ except in the case $x=0$.

Note that $5040=7!$. So the approximation obtained by stopping at the term $\frac{x^5}{5!}$ has error with absolute value $\lt \frac{1}{5040}$.

Remark: For $\sin x$, and $|x|\le 1$, you can get the same estimate of the error by using the Lagrange form of the remainder. Since it is likely that you will be asked similar questions where we do not have an alternating series, it is useful to know how to use the Lagrange form of the remainder,

$\endgroup$
0
$\begingroup$

Hint: The error of an alternating series $\sum_k a_k $ can be bounded as

$$ |R_n|\leq a_{n+1}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.