Taylor Series Approximation

[Idealistic]

Provide a polynomial p(x) such that abs(sin(x) - p(X)) < 10[sup:1dntfj8r]-8[/sup:1dntfj8r] for -pi/4 < x < pi/4

Don't think the question is asking for integral test errors; given that the taylor series for sin(x) is:

sum[x[sup:1dntfj8r](2n + 1)[/sup:1dntfj8r]/(2n + 1)]

How do I ensure that my polynomial is accurate (less than 10^-8) on the interval (-pi/4, pi/4)?

 

[Deleted member 4993]

Provide a polynomial p(x) such that abs(sin(x) - p(X)) < 10[sup:2jesa4u2]-8[/sup:2jesa4u2] for -pi/4 < x < pi/4

Don't think the question is asking for integral test errors; given that the taylor series for sin(x) is:

sum[x[sup:2jesa4u2](2n + 1)[/sup:2jesa4u2]/(2n + 1)]

How do I ensure that my polynomial is accurate (less than 10^-8) on the interval (-pi/4, pi/4)?

You need to look at the residual (remainder) for Taylor's series for sum up to n-terms. check out your text-book or:

http://en.wikipedia.org/wiki/Taylor%27s_theorem

 

[Idealistic]

So my a = 0?, and how do I ensure the interval is from (-pi/4, pi/4)?

 

[Idealistic]

am I setting 10^-8 = f(x) - sum[x(2n + 1)/(2n + 1)]

 

[daon]

Try the althernating series approximation:

\(\displaystyle |R_n| \le |S-S_n| \le b_{n+1}\)

where \(\displaystyle b_n\) is, in this case:

\(\displaystyle \frac{x^{2n+1}}{(2n+1)!}\)