1. ## 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?

2. Originally Posted by casanova
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?

3. Originally Posted by casanova
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 ?

4. 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...

5. 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?

6. Are you sure that's the right "hint"?

7. 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 .

8. I see it now *hangs head in shame* so easy!!!! You don't even really need Taylor's theroem if you use that hint.