# Third Order Taylor Polynomial

• Apr 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!
• Apr 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?

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

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

Quote:

also another question, the series 4^n+1/3^n is geometric correct? thanks!
$\displaystyle 4^n + \frac{1}{3^n}$ is the sum of two geometric series.
• Apr 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
• Apr 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 $\displaystyle f^4(x)$ over the interval $\displaystyle 2 \leq x \leq 2.1$

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

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

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

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