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# 数学代写|实分析代写Real Analysis代考|Math444 The Problems with this Proof

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## 数学代写|实分析代写Real Analysis代考|The Problems with this Proof

This is an ingenious proof that demonstrates a profound understanding of the derivative, but it would not pass muster today. We can prove something stronger than Cauchy’s theorem. Cauchy required that $f$ be differentiable at every point in the interval $[a, b]$. This is necessary if we want to get all of the inequalities of (3.24). As we shall see, Bonnet’s proof works under the slightly weaker assumption that we have differentiability on the open interval $(a, b)$ and continuity on the closed interval $[a, b]$. For example, Cauchy’s proof would not apply to the function $x \sin (1 / x)$ on $[0,1]$ (see exercise 3.1 .12 of section 3.1 and exercise 3.2.1 of this section). Bonnet’s would. Furthermore, Cauchy only proves that $c$ lies somewhere in the interval $[a, b]$. Bonnet’s proof shows that it must lie strictly between $a$ and $b$.

But these are quibbles. The question is not whether we can improve on Cauchy’s statement of the mean value theorem, but whether his proof is valid. There are two questionable assertions in his proof. The first is that given the error bound $\epsilon$ we can find a $\delta$ that works over the entire interval $[a, b]$. It is true that at $x_0$ we can find a $\delta$ so that if $\left|x_1-x_0\right|<\delta$, then
$$f^{\prime}\left(x_0\right)-\epsilon<\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}<f^{\prime}\left(x_0\right)+\epsilon,$$
but the inequalities of (3.24) assume that the same $\delta$ can be used at $x_1$ and $x_2$ and $x_3 \ldots$ all the way to $x_{n-1}$. This is not always so (see exercise 3.2 .2 for a counterexample).

## 数学代写|实分析代写Real Analysis代考|Cauchy’s Second Attempt

Cauchy gave another proof of the mean value theorem in an appendix to Résumé des Leçons. He actually proves a far more powerful result. We shall state it in the form in which we will eventually prove it.

Theorem 3.2 (Generalized Mean Value Theorem). If $f$ and $F$ are both continuous at every point of $[a, b]$ and differentiable at every point in the open interval $(a, b)$ and if $F^{\prime}$ is never zero in this interval, then
$$\frac{f(b)-f(a)}{F(b)-F(a)}=\frac{f^{\prime}(c)}{F^{\prime}(c)}$$
for at least one point $c, a<c<b$.
We note that if $F(x)=x$, then this becomes the ordinary mean value theorem.
We begin by defining $g(x)=f(x)-f(a)$ and $G(x)=F(x)-F(a)$ so that $g(a)=0$, $g^{\prime}(x)=f^{\prime}(x), G(a)=0$, and $G^{\prime}(x)=F^{\prime}(x)$. We consider the function defined by
$$\frac{f^{\prime}(x)}{F^{\prime}(x)}=\frac{g^{\prime}(x)}{G^{\prime}(x)}$$
and let $A$ be its minimal value over $[a, b], B$ its maximal value:
$$A \leq \frac{g^{\prime}(x)}{G^{\prime}(x)} \leq B$$

.

## 数学代写|实分析代写Real Analysis代考|The Problems with this Proof

$$f^{\prime}\left(x_0\right)-\epsilon<\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}<f^{\prime}\left(x_0\right)+\epsilon$$

## 数学代写|实分析代写Real Analysis代考|Cauchy’s Second Attempt

$$\frac{f(b)-f(a)}{F(b)-F(a)}=\frac{f^{\prime}(c)}{F^{\prime}(c)}$$

$$\frac{f^{\prime}(x)}{F^{\prime}(x)}=\frac{g^{\prime}(x)}{G^{\prime}(x)}$$

$$A \leq \frac{g^{\prime}(x)}{G^{\prime}(x)} \leq B$$

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