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Math Help - Taylor Polynomial

  1. #1
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    Taylor Polynomial

    My professor tried to explain Taylor Polynomials and I'm having trouble with notation...

    He gives us the following exercise...

    Let f(x)=\frac{1}{1-x}. Let x_0=0.
    1) Obtain an expression for R_{1}f.
    2) Find the value of c when x=3/4.

    He tells us that f(x)=T_{n}f(x)+R_{n}f(x) where R_{n}f(x)=\frac{f^{n+1}(c)}{(n+1)!}(x-x_0)^{n+1} for some c between x_0 and x.

    I feel like this should be very simple but he didn't do any examples in class so the notation is confusing me. Any help would be appreciated.
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  2. #2
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    Consider Taylor's theorem: under certain conditions, a function f(x) can be represented as  f(x) = f(x_0) + \frac{f'(x_0)}{1!}(x - x_0) + \frac{f^{(2)}(x_0)}{2!}(x - x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n + R_n(x) for some function R_n(x). Your professor's notation for f(x_0) + \frac{f'(x_0)}{1!}(x - x_0) + \frac{f^{(2)}(x_0)}{2!}(x - x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n is T_nf(x), and the notation for R_n(x) is R_nf(x). (In these notations, T_n and R_n are viewed as operators that take functions and return functions.) Now, the theorem above is, of course, obvious if nothing more is stated about R_n(x). The theorem also states that R_n(x) can be represented as \frac{f^{n+1}(c)}{(n+1)!}(x-x_0)^{n+1} for some c between x and x_0.

    Note that 1/(1-x) = 1 + x + x^2 +\dots for \lvert x\rvert<1.
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  3. #3
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    So if I wanted to find an expression for R_{1}f then I would plug in n=1?

    So I would have R_{1}f=\frac{f^{2}(c)}{2!}(x-0)^2 ?
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  4. #4
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    Quote Originally Posted by zebra2147 View Post
    So if I wanted to find an expression for R_{1}f then I would plug in n=1?
    Yes.

    So I would have R_{1}f=\frac{f^{2}(c)}{2!}(x-0)^2 ?
    Yes. Actually, I would reformulate Taylor's theorem like this. Let R_nf(x) be f(x) - T_nf(x). Then for every x there exists a c such that R_nf(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}. So, by definition, R_1(1/(1-x))=1/(1-x)-1-x. Taylor's theorem claims that \displaystyle R_1(1/(1-x))=\frac{1}{2}\left(\frac{1}{1-x}\right)''(c)\cdot x^2. You have to find c when x = 3/4.
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