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Let

$$f(x) = a_{n+1}x^{n+1} + a_nx^n + ... + a_1x + a_0$$ be a polynomial of degree $n+1$ defined for all $x\in\mathbb{R}$. Show that:

(a)The derivative of order $n+1$ of $f$ is equal to: $$f^{(n+1)}(x) = (n+1)!a_{n+1}$$

First we deal with the monomial $x^k, k\in[0,n]$, showing that the (n+1)th derivative this is zero. So we proceed via induction and let; $$P(n):=\text{ the } (n+1)\text{th derivative of } x^k, k \in[0,n) \text{ is } 0$$

$P(1)$

If $k = 0$ then $\frac{d^2(1)}{dx^2} = 0$

and so $P(1)$ holds.

Suppose that $P(n)$ for some $n\in\mathbb{N}$, we need to show the truth of $P(n+1)$

If $k\in[0,n)$ then the statement is vacuously true by the inductive hypothesis. If $k = n$, we have that the $n$th derivative of $x^{n}$ to be

$$\frac{d^n{(x^{n})}}{dx^n}=n!$$

on taking derivatives of this expression, we obtain

$$\frac{d^{n+1}{(x^{n})}}{dx^{n+1}}=0$$

and taking derivatives again, as we require to show that the $((n+1)+1)th$ derivative is zero, we obtain

$$\frac{d^{n+2}{(x^{n})}}{dx^{n+2}}=0$$ since $n!$ is constant.

Hence by mathematical induction we have shown the truth of $P$ for all $n$.

From the discussion above we note that $$\frac{d^{n+1}{(x^{n+1})}}{dx^{n+1}}=(n+1)!=\frac{f^{(n+1)}(x)}{a_{n+1}}$$

multiplying through we obtain the result.

(b) Deduce that, for any polynomial $p_n$ interpolating f at any distinct points $x_0<x_1<...<x_n$,the following identity holds:

$$f(x)-p_n(x)=a_{n+1}(x-x_0)...(x-x_n)$$

By theorem 3.3(Interpolation error theorem) there exists $\xi(x) \in (x_0,x_n)$ such that $$ f(x)-p_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)...(x-x_n)$$

Since out expression from (a) is independent of $x$, we can write

$$f(x)-p_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)...(x-x_n)=a_{n+1}(x-x_0)...(x-x_n)$$

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  • $\begingroup$ For the first part, use linearity to work one term at a time: show that all the terms except the leading one vanish eventually, and then you just need to know how to handle $x^{n+1}$. $\endgroup$ – Ian Nov 2 '15 at 13:47
  • $\begingroup$ I'm really sorry, what do you mean by "use linearity"? $\endgroup$ – user197848 Nov 2 '15 at 14:01
  • $\begingroup$ The derivative operator is linear, so the $(n+1)$th derivative operator is also linear. That means that if you are differentiating a polynomial a bunch of times, it is enough to differentiate the monomials and then add them back together with the coefficients. $\endgroup$ – Ian Nov 2 '15 at 14:02
  • $\begingroup$ Ah i see, so I could say the $i$th monomial, $i\in[0,n]$ , after $i$ differentiations will be constant and hence after $i+1$ differentiation's will be $0$. concerning the $n+1$th case we know that after the $i$th each differentiation the coefficient of the $x^{n+1}$ term will be multiplied by $n+2-i$ since $i\in[0,n]$, and so after n+1 differentiations we have $(n+1).....(n+2-i).....(1)x^{0}a_{n+1}$. Is that along the correct lines? Its a bit rough, but I'd have to treat it as an inductive proof right? $\endgroup$ – user197848 Nov 2 '15 at 14:27
  • $\begingroup$ Strictly speaking you need induction, yes: you want to argue that the $(n+1)$th derivative of $x^k$ is $0$ for $k<n+1$ and $(n+1)!$ for $k=n+1$. Then the result will follow. $\endgroup$ – Ian Nov 2 '15 at 14:28
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In the first part, argue that the $(n+1)$th derivative of $x^k$ is $0$ for $k<n+1$ and $(n+1)!$ for $k=n+1$. Then use linearity of the derivative to finish the proof.

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  • $\begingroup$ Hi Ian, I have edited by question to include to include the inductive proof. If you have time, could you take a little look. Thanks regardless! :) $\endgroup$ – user197848 Nov 2 '15 at 15:51

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