1. ## Taylor Polynomial

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

He gives us the following exercise...

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

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

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.

2. Consider Taylor's theorem: under certain conditions, a function $\displaystyle f(x)$ can be represented as $\displaystyle f(x) = f(x_0) + \frac{f'(x_0)}{1!}(x - x_0) + \frac{f^{(2)}(x_0)}{2!}(x - x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n + R_n(x)$ for some function $\displaystyle R_n(x)$. Your professor's notation for $\displaystyle f(x_0) + \frac{f'(x_0)}{1!}(x - x_0) + \frac{f^{(2)}(x_0)}{2!}(x - x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n$ is $\displaystyle T_nf(x)$, and the notation for $\displaystyle R_n(x)$ is $\displaystyle R_nf(x)$. (In these notations, $\displaystyle T_n$ and $\displaystyle R_n$ are viewed as operators that take functions and return functions.) Now, the theorem above is, of course, obvious if nothing more is stated about $\displaystyle R_n(x)$. The theorem also states that $\displaystyle R_n(x)$ can be represented as $\displaystyle \frac{f^{n+1}(c)}{(n+1)!}(x-x_0)^{n+1}$ for some $\displaystyle c$ between $\displaystyle x$ and $\displaystyle x_0$.

Note that $\displaystyle 1/(1-x) = 1 + x + x^2 +\dots$ for $\displaystyle \lvert x\rvert<1$.

3. So if I wanted to find an expression for $\displaystyle R_{1}f$ then I would plug in $\displaystyle n=1$?

So I would have $\displaystyle R_{1}f=\frac{f^{2}(c)}{2!}(x-0)^2$ ?

4. Originally Posted by zebra2147
So if I wanted to find an expression for $\displaystyle R_{1}f$ then I would plug in $\displaystyle n=1$?
Yes.

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