0
$\begingroup$

Let $f(x)=\frac{\sin(x)}{x}$ when $x\neq 0$ and $f(x)=1$ when $x=0$. Starting with the Taylor polynomial of degree $2n+1$ for $\sin(x)$ and the estimate for the remainder term, show that

$f(x)=(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots +(-1)^n \frac{x^{2n}}{(2n+1)!})+R_{2n,0,f}(x)$ where $|R_{2n,0,f}(x)|\leq \frac{|x|^{2n+1}}{(2n+2)!}$ and use this to conclude that $\int_0^1 f \approx \int_0^1 (1-\frac{x^2}{3!}+\frac{x^4}{5!})dx = \frac{1703}{1800}\approx .946$ with an error of less than $10^{-3}$.

The first part of this seems pretty straightforward unless I'm missing something with the piecewise nature of $f$. The second statement is not immediately clear to me, though.

$\endgroup$
0
$\begingroup$

Basically you estimate $ f (x) $ with the first 3 terms of it's expansion. There will be an error involved, whose value is given in your question. Since we are taking the expansion up to the $ n=2$ term, our remainder is

$ R \leq \frac {x^5}{6!}$

To estimate$ \int_0^1 f (x) dx$ we take the integral of the first three terms of the expansion (which is easily computed). The error involved then is

$\displaystyle\int_0^1 \frac {x^5}{6!} = \frac {1}{6(6!)} = \frac {1}{4320} \leq 10^{-3}$ So you have your second part

| cite | improve this answer | |
$\endgroup$

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.