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Thread: Third Order Taylor Polynomial

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    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!
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    Quote Originally Posted by TAG16 View Post
    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

    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.
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    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
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    Quote Originally Posted by TAG16 View Post
    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!} $
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