Function Approximation

**larson** · March 5th 2008, 07:08 PM

I'm having trouble with the following question:

Let f(x) = sinx and P5(x) be the fifth-order Maclaurin polynomial for f.

a) What does Theorem 2 imply about the maximum approximation error committed by P5 over the interval [-2,2]?

b) What is the maximum actual approximation error committed by P5 over the interval [-2,2]?

Any help would be greatly appreciated

.

**ThePerfectHacker** · March 5th 2008, 07:29 PM

Let $P_5(x)$ be the 5th degree Taylor polynomial, and $f(x) = \sin x$ . Define $R_5(x) = f(x) - P_5(x)$ then if $x_0$ is any non-zero point on $(-2,2)$ it means by Lagrange's remainder there exists $y$ between $0$ and $x_0$ so that $R_5(x_0) = \frac{f^{(6)}(y)}{6!} y^{6}$ . Now $f^{(6)}(y) = -\sin y$ . This means $|R_5(x_0)| = \left| \frac{-\sin y}{6!} y^6 \right| \leq \frac{2^6}{6!} = .0\bar 8$ , but maybe we can improve this estimate. Let us use Cauchy's remainder, $R_5(x_0) = \frac{f^{(6)}(y)}{6!}(x_0 - y)^n x_0$ , note $|x_0 - y|\leq 1$ and $|x_0|\leq 1$ thus, $|R_5(x_0)| \leq \frac{1}{6!} = .0013\bar 8$ .

**TheEmptySet** · March 5th 2008, 07:33 PM

Originally Posted by larson

I'm having trouble with the following question:

Let f(x) = sinx and P5(x) be the fifth-order Maclaurin polynomial for f.

a) What does Theorem 2 imply about the maximum approximation error committed by P5 over the interval [-2,2]?

b) What is the maximum actual approximation error committed by P5 over the interval [-2,2]?

Any help would be greatly appreciated

.

I not sure what theorem 2 is in your book, but I would guess it would be Taylor's Inequality...

if
$|f^{n+1}(x)| \le M$ for $|x-a|\le d$ , then the remainder $R_n(x)$ of the Taylor series satisfies the inequality

$|R_n(x)|\le\frac{M}{(n+1)!}|x-a^{n+1}|$

Since the 6th derivative of Sin(x) is -Sin(x)

Its max on [-2,2] is 1 so M =1. so in the formula a=0 because it is a Maclaurin series n=5 because it is 5th degree taylor poly.

Plugging all that in gives us

$|R_n(x)| \le \frac{1}{(5+1)!}2^{5+1}=\frac{64}{720}=\frac{4}{45 }$

The alternating series estimation could also be used and is much easier as well.
simplifying

**larson** · March 5th 2008, 08:23 PM

Thank you so so much.

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