# Taylor Polynomial (Approximation Error)

• Mar 18th 2008, 05:27 PM
larson
Taylor Polynomial (Approximation Error)
Hey I'm having trouble with the 2nd part of this question:

Find the fifth-order Taylor polynomial, then find an interval centered at x = x(o) in which the approximation error $|f(x) - P5(x)|$ is less than 0.01.

$f(x) = \sqrt {x} : x(o) = 1$

So I found the fifth-order taylor polynomial which is:

$P5(x) = 1 + \frac {1}{2}(x-1) - \frac {1}{8}(x-1)^2 + \frac {1}{16}(x-1)^3 - \frac {5}{128}(x-1)^4 + \frac {7}{256}(x-1)^5$

Now how do I find the approximation error? Thanks in advance.
• Mar 18th 2008, 06:23 PM
TheEmptySet
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

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

so $\frac{d^6f}{dx^6}=-\frac{945}{64x^{11/2}}$

The 6th derivative has it max at x=1 for all values greater than 1.(it is unbounded as we approch zero from the right) $|-\frac{945}{64}|=M$

we want the error to be less than 0.01

we need to solve

$\frac{945}{64 \cdot 6!}|x-1|^6 \le 0.01$

$|x-1|^{6} \le 0.0001190$

$|x-1| \le .2218$
• Mar 18th 2008, 07:02 PM
larson
Quote:

Originally Posted by TheEmptySet
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

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

so $\frac{d^6f}{dx^6}=-\frac{945}{64x^{11/2}}$

The 6th derivative has it max at x=1 for all values greater than 1.(it is unbounded as we approch zero from the right) $|-\frac{945}{64}|=M$

we want the error to be less than 0.01

we need to solve

$\frac{945}{6!}|x-1|^6 \le 0.01$

$|x-1|^{6} \le 0.007619$

$|x-1| \le .4436$

Didn't you forget your 64 in the denominator? Just curious, but other than that I get it.
• Mar 18th 2008, 07:07 PM
TheEmptySet
Quote:

Originally Posted by larson
Didn't you forget your 64 in the denominator? Just curious, but other than that I get it.

Yes I did. I have been off today.

$
\frac{945}{64 \cdot 6!}|x-1|^6 \le 0.01
$

Thanks

I will fix the above post
• Mar 18th 2008, 07:16 PM
larson
Quote:

Originally Posted by TheEmptySet
$\frac{945}{64 \cdot 6!}|x-1|^6 \le 0.01$

$|x-1|^{6} \le 0.0001190$

$|x-1| \le .2218$

I'm having trouble now with this...

When you do $\frac{945}{64 \cdot6!}$ you get...

$\frac{945}{46080}$ which then equals $.0205078125$

Then you divide both sides by that so wouldn't you get... $|x-1|^6 \le \frac{0.01}{.0205078125}$ which means... $|x-1|^6 \le .487619$ right?

Sorry if I I'm just not understanding it correctly...
• Mar 18th 2008, 07:29 PM
TheEmptySet
$\frac{945}{64 \cdot 6!}|x-1|^6 \le .01$

$|x-1|^6 \le \frac{.01 \cdot 64 \cdot 6!}{945}=\frac{460.8}{945}$

$|x-1|^6 \le 0.4876 \iff |x-1| \le .8872$

You are correct I just can't read or type today.

Sorry(Worried)
• Mar 18th 2008, 07:38 PM
larson
Quote:

Originally Posted by TheEmptySet
$\frac{945}{64 \cdot 6!}|x-1|^6 \le .01$

$|x-1|^6 \le \frac{.01 \cdot 64 \cdot 6!}{945}=\frac{460.8}{945}$

$|x-1|^6 \le 0.4876 \iff |x-1| \le .8872$

You are correct I just can't read or type today.

Sorry(Worried)