Third Order Taylor Polynomial

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April 16th 2009, 01:29 PM
TAG16

Third Order Taylor Polynomial

If the function f had the following properties: f(2)=1 f'(2)=-3 f''(2)=2 and f'''(x)= (1+x^3)^1/2 for all x> 1 , then use a third order Taylor polynomial to approximate f(2.1). How would you solve this?

also another question, the series 4^n+1/3^n is geometric correct? thanks!
April 16th 2009, 02:00 PM
skeeter

Quote:

Originally Posted by TAG16

If the function f had the following properties: f(2)=1 f'(2)=-3 f''(2)=2 and f'''(x)= (1+x^3)^1/2 for all x> 1 , then use a third order Taylor polynomial to approximate f(2.1). How would you solve this?

$f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!}$

$a = 2$ , $x = 2.1$ ... plug and chug

Quote:

also another question, the series 4^n+1/3^n is geometric correct? thanks!

$4^n + \frac{1}{3^n}$ is the sum of two geometric series.
April 16th 2009, 04:28 PM
TAG16

How would you use Taylor's inequality ( l Rn l < M l x-a l^n+1/(n+1)! ) to find the upper bound for the error l Rn l ? I'm just not sure how to find M
April 16th 2009, 04:53 PM
skeeter

Quote:

Originally Posted by TAG16

How would you use Taylor's inequality ( l Rn l < M l x-a l^n+1/(n+1)! ) to find the upper bound for the error l Rn l ? I'm just not sure how to find M

M will be the maximum value of $f^4(x)$ over the interval $2 \leq x \leq 2.1$

$f'''(x) = (1+x^3)^{\frac{1}{2}}$

$f^4(x) = \frac{3x^2}{2(1+x^3)^{\frac{1}{2}}}$

max will occur when $x = 2.1$ since $f^4(x)$ is increasing for all $x > 0$

$|R| \leq \frac{f^4(2.1)(2.1 - 2)^4}{4!}$