- #1

zenterix

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- TL;DR Summary
- In Chapter 20, "Approximation by Polynomial Functions", there is a lemma (used to prove Taylor's Theorem right after), the proof of which includes a step that I would like to understand better.

In Chapter 20 of Spivak's

For ##n=0## we have

$$\frac{R(x)}{x-a}=\frac{R(x)-R(a)}{x-a}=R'(t)$$

for some ##t\in (a,x)##, which is the MVT applied to ##R## on ##[a,x]##.

Now assume as our induction hypothesis that ##(1)## is true for some ##k##.

First, let's apply Cauchy MVT to the functions ##R(x)## and ##g(y)=(y-a)^{k+2}##.

$$\frac{R(x)-R(a)}{g(x)-g(a)}=\frac{R'(z)}{g'(z)}$$

for some ##z\in (a,x)##. Then

$$\frac{R(x)}{g(x)}=\frac{R(x)}{(x-a)^{k+2}}=\frac{R'(z)}{(k+2)(z-a)^{k+1}}=\frac{1}{k+2}\frac{R'(z)}{(z-a)^{k+1}}$$

At this point, we are to apply the induction hypothesis to ##R'## on ##[a,z]##. In the induction hypothesis, ##R## is assumed to be ##(n+1)##-times differentiable. Doesn't this mean that ##R'## is only ##n##-times differentiable?

How do we know that ##(R')^{n+1}## is defined in ##(2)##?

*Calculus*is the lemma shown below (used afterward to prove Taylor's Theorem). My question is about a step in the proof of this lemma.Lemma:Suppose that the function ##R## is ##(n+1)##-times differentiable on ##[a,b]## and

$$R^{(k)}(a)=0, \text{ for } k=0,1,2,...,n$$

Then for any ##x## in ##(a,b]## we have

$$\frac{R(x)}{(x-a)^{n+1}}=\frac{R^{(n+1)}(t)}{(n+1)!}, \text{ for

some } t \text{ in } (a,x)\tag{1}$$

**Here is the proof as it appears in the book**For ##n=0## this is just the Mean Value Theorem, and we will prove the

theorem for all ##n## by induction on ##n##. To do this we use the Cauchy

Mean Value Theorem to write

$$\frac{R(x)}{(x-a)^{n+2}}=\frac{R'(z)}{(n+2)(z-a)^{n+1}}=\frac{1}{n+2}\frac{R'(z)}{(z-a)^{n+1}},

\text{ for some } z \text{ in } (a,x)$$

and then apply the induction hypothesis to ##R'##on the interval

$[a,z]$ to get

$$\frac{R(x)}{(x-a)^{n+2}}=\frac{1}{n+2}\frac{(R')^{(n+1)}(t)}{(n+1)!},

\text{ for some } t \text{ in } (a,z)\tag{2}$$

$$=\frac{R^{(n+2)}(t)}{(n+2)!}$$

**My question is:**how do we know that ##(R')^{n+1}## is defined in ##(2)##?**Let me try to go through the proof in more steps.**For ##n=0## we have

$$\frac{R(x)}{x-a}=\frac{R(x)-R(a)}{x-a}=R'(t)$$

for some ##t\in (a,x)##, which is the MVT applied to ##R## on ##[a,x]##.

Now assume as our induction hypothesis that ##(1)## is true for some ##k##.

First, let's apply Cauchy MVT to the functions ##R(x)## and ##g(y)=(y-a)^{k+2}##.

$$\frac{R(x)-R(a)}{g(x)-g(a)}=\frac{R'(z)}{g'(z)}$$

for some ##z\in (a,x)##. Then

$$\frac{R(x)}{g(x)}=\frac{R(x)}{(x-a)^{k+2}}=\frac{R'(z)}{(k+2)(z-a)^{k+1}}=\frac{1}{k+2}\frac{R'(z)}{(z-a)^{k+1}}$$

At this point, we are to apply the induction hypothesis to ##R'## on ##[a,z]##. In the induction hypothesis, ##R## is assumed to be ##(n+1)##-times differentiable. Doesn't this mean that ##R'## is only ##n##-times differentiable?

How do we know that ##(R')^{n+1}## is defined in ##(2)##?