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Thread: Taylor's Thm

  1. #1
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    Taylor's Thm

    Suppose that $\displaystyle f^{(n+1)}$ exists on $\displaystyle (a, b)$ and $\displaystyle x_0, x_1, ..., x_n \in (a, b)$ and p is the polynomail of degree $\displaystyle \leq n$ such that $\displaystyle p(x_i)=f(x_i)$. I want to show that for $\displaystyle x\in (a, b)$:

    $\displaystyle f(x)=p(x)+\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)(x-x_1)...(x-x_n)$

    for $\displaystyle c_x \in(a, b)$

    We are just starting Taylor's Theorem but I'm not sure how this works. Any advice?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by casanova View Post
    Suppose that $\displaystyle f^{(n+1)}$ exists on $\displaystyle (a, b)$ and $\displaystyle x_0, x_1, ..., x_n \in (a, b)$ and p is the polynomail of degree $\displaystyle \leq n$ such that $\displaystyle p(x_i)=f(x_i)$. I want to show that for $\displaystyle x\in (a, b)$:

    $\displaystyle f(x)=p(x)+\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)(x-x_1)...(x-x_n)$

    for $\displaystyle c_x \in(a, b)$

    We are just starting Taylor's Theorem but I'm not sure how this works. Any advice?
    What do you think?
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  3. #3
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    Quote Originally Posted by casanova View Post
    Suppose that $\displaystyle f^{(n+1)}$ exists on $\displaystyle (a, b)$ and $\displaystyle x_0, x_1, ..., x_n \in (a, b)$ and p is the polynomail of degree $\displaystyle \leq n$ such that $\displaystyle p(x_i)=f(x_i)$. I want to show that for $\displaystyle x\in (a, b)$:

    $\displaystyle f(x)=p(x)+\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)(x-x_1)...(x-x_n)$

    for $\displaystyle c_x \in(a, b)$

    We are just starting Taylor's Theorem but I'm not sure how this works. Any advice?
    Note that $\displaystyle (x-x_0)(x-x_1)...(x-x_n)$ is a polynomial of degree $\displaystyle n+1 $ while $\displaystyle p(x) $ , its degree doesn't exceed $\displaystyle n$ .Therefore, the leading coefficient of $\displaystyle (x-x_0)(x-x_1)...(x-x_n)$ is just the coefficient of $\displaystyle x^{n+1} $ in Taylor's expansion for $\displaystyle f(x) $ . But has the degree of $\displaystyle f(x) $ been given ?
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  4. #4
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    The only information I have about f(x) is given above. I was trying to introduce a new polynomial that did $\displaystyle f(x)-p(x)$ to try to get the rest of the right hand side. But I kept getting stuck because I don't know anything about f(x) or p(x). I knew I could say:

    $\displaystyle f(x) = f(x_0) +f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!} (x-x_0)^2+...$
    $\displaystyle +\frac{f^{(n)}(x_0)}{n!} (x-x_0)^n+\frac{f^{(n+1)}(c)}{(n+1)!} (x-x_0)^{n+1}$

    But all I know about p(x) is that $\displaystyle p(x_i)=f(x_i)$ so I thought if I subtracted the 2 in the new polynomial, I might get close because of this equality... but it doesn't end up giving me anything that looks like what I want to show because I don't know about the derivatives of p...
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  5. #5
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    I just realized theres a hint in the back of the book that says to do the following:

    let x be fixed and distinct from $\displaystyle x_0, ..., x_n$ and consider function:

    $\displaystyle g(y) = f(y)-p(y)-\frac{K}{(n+1)!}(y-x_0)(y-x_1)...(y-x_n)$

    where we choose K so that $\displaystyle g(x) = 0$

    then use Rolles thm to show $\displaystyle K = f^{(n+1)}(c)$ for $\displaystyle c \in (a, b)$

    Rolles???? How does that relate at all? this confused me more than helped i think? any ideas?
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  6. #6
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    Are you sure that's the right "hint"?
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  7. #7
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    Oh , I missed a point , you said $\displaystyle c_x \in (a,b)$ , that means the choice of $\displaystyle c $ depends on $\displaystyle x $ , you are correct ! Use Taylor's Theorem to show that .
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  8. #8
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    I see it now *hangs head in shame* so easy!!!! You don't even really need Taylor's theroem if you use that hint.
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