A Cauchy-L'Hopital-Taylor hybrid theorem

Last week I was teaching the standard theorems of real analysis (single variable): Rolle's Theorem, the Mean Value Theorem (MVT), Cauchy's Mean Value Theorem, L'Hôpital's rule, and Taylor's Theorem. As I was short of time, I didn't manage to give full proofs of everything live (but they have complete proofs in the printed notes). I did however have a think about the relationships between these results, and what I might say about them.

Along the way, I noticed that there is a particular hybrid Cauchy-L'Hôpital-Taylor result which I don't remember seeing before (though no doubt it is well known).

Let me start by recalling some of the standard statements.

For the rest of this post, $a$ and $b$ are real numbers with $a<b,$ and $f$ and $g$ are real-valued functions on $[a,b]$ which are both continuous on $[a,b]$ and differentiable on $(a,b).$

Rolle's Theorem

Suppose that $f(a)=f(b)$. Then there exists $\xi \in (a,b)$ such that $f'(\xi)=0\,.$

Comment: We proved this using a theorem often credited to Fermat about derivatives (when they exist) at local extrema of a function in the interior of an interval.

Mean Value Theorem (MVT)

Note: here there is no assumption on $f(a)$ and $f(b)$.

There exists $\xi \in (a,b)$ such that

$$f(b)-f(a) = (b-a)f'(\xi)\,.$$

Equivalently, there exists $\xi \in (a,b)$ such that

$$f'(\xi) =\frac{f(b)-f(a)}{b-a}\,.$$

Comment: This says that the gradient of $f$ at $\xi$ is equal to the average gradient of $f$ over the whole interval $[a,b].$ Of course, Rolle's Theorem is the special case where this average gradient is $0.$ However, we use Rolle's Theorem to prove the MVT: subtracting of a suitable linear polynomial from $f$ gives a new function which satisfies the conditions of Rolle's Theorem. Any linear polynomial with gradient $\frac{f(b)-f(a)}{b-a}$ will do.

To help prove L'Hôpital's rule we need a more general version of the MVT. This is the first time we bring our second function $g$ into the picture. Remember that both $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$.

Cauchy's Mean Value Theorem (CMVT)

There exists $\xi \in (a,b)$ such that

$$(f(b)-f(a))g'(\xi) = (g(b)-g(a))f'(\xi)\,.$$

Note: this is the same $\xi$ for $f$ and $g$, so that we can't just apply the usual MVT separately to $f$ and $g$.

When proving L'Hôpital's rule, the main case of interest is when we know that $g'(t)\neq 0$ for all $t \in (a,b)$ and (hence) that $g(b)\neq g(a).$ In this case, the point $\xi$ above gives us

$$\frac{f'(\xi)}{g'(\xi)} = \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{\frac{f(b)-f(a)}{b-a}}{\frac{g(b)-g(a)}{b-a}}\,.$$

This says that the ratio of the gradients of $f$ and $g$ at $\xi$ is equal to the ratio of the average gradients of $f$ and $g$ over the interval $[a,b].$ 

If, in addition to the above assumptions on $g$ we also have $f(a)=f(b)=0,$ then we have $f(b)/g(b) = f'(\xi)/g'(\xi),$ and this helps with the proof one of the main versions of L'Hôpital's rule. However, I don't want to discuss all the different forms of L'Hôpital's rule here.

Note that the usual MVT is just the special case of the CMVT where $g(x)=x$ (or $x+c$ for any constant $c$.)

At this point, we usually prove some version of Taylor's Theorem with remainder (there are quite a few versions). But I want to look at a more CMVT-style version, based on what I will call a Hybrid Lemma.

Hybrid Lemma

Suppose now (in addition to our initial assumptions) that $n\in\{0,1,2,\dots\},$ and that $f$ and $g$ are $n+1$ times differentiable on $(a,b).$

Let $x_0 \in (a,b),$ and suppose that, for all $k \in \{0,1,2,\dots,n\},$ we have $f^{(k)}(x_0)=g^{(k)}(x_0)=0.$ Then there exist $\xi_a \in (a,x_0)$ and $\xi_b \in (x_0,b)$ such that

$(1)\Jdisplay f(a) g^{(n+1)}(\xi_a) = g(a) f^{(n+1)}(\xi_a)$

and

$(2)\Jdisplay f(b) g^{(n+1)}(\xi_b) = g(b) f^{(n+1)}(\xi_b)\,.$

Comments:

Since $f(x_0)=g(x_0)=0$, the case $n=0$ is immediate from CMVT on the intervals $[a,x_0]$ and $[x_0,b].$

The main case of interest for us is when $g^{(n+1)}(t) \neq 0$ for all $t \in (a,b)\setminus\{x_0\}$ and (hence) $g(a) \neq 0$ and $g(b)\neq 0.$ In this case the conclusion is that

$(3)\Jdisplay \frac{f(a)}{g(a)} =\frac{f^{(n+1)}(\xi_a)}{g^{(n+1)}(\xi_a)}$

and

$(4)\Jdisplay \frac{f(b)}{g(b)} =\frac{f^{(n+1)}(\xi_b)}{g^{(n+1)}(\xi_b)}\,.$

Because of symmetry, there is no real difference between the results for $\xi_a$ and for $\xi_b,$ so we will focus on $\xi_b.$

Let's first look at the case we are most interested in, where $g^{(n+1)}$ has no zeros in $(a,b)\setminus\{x_0\}.$ Since $g^{(n)}(x_0)=0,$ Rolle's Theorem then tells us that $g^{(n)}$ also has no zeros in $(a,b)\setminus\{x_0\}.$ Repeating this argument, we see that for all $k \in \{0,1,2,\dots,n\},$ $g^{(k)}$ has no zeros in $(a,b)\setminus\{x_0\}.$ Moreover $g(a)\neq 0$ and $g(b) \neq 0\,.$ 

[You can also write this out as a more formal proof by induction, but I think that it is clear enough in this form.]

We now prove the existence of $\xi_b$ satisfying $(4)$.

We first apply CMVT on the interval $[x_0,b].$ This tells us that there exists $\xi_1 \in (x_0,b)$ with

$$\frac{f(b)}{g(b)} =\frac{f'(\xi_1)}{g'(\xi_1)}\,.$$

We can now proceed by induction, or by simply repeatedly applying CMVT. By CMVT, applied to $f'$ and $g'$ on $[x_0,\xi_1]$, there is a $\xi_2\in (x_0,\xi_1)$  with 

$$\frac{f(b)}{g(b)} =\frac{f'(\xi_1)}{g'(\xi_1)} = \frac{f''(\xi_2)}{g''(\xi_2)}\,.$$

Repeated application of CMVT gives us $b>\xi_1>\xi_2>\cdots >\xi_{n+1} > x_0$ with

$$\frac{f(b)}{g(b)} =\frac{f^{(1)}(\xi_1)}{g^{(1)}(\xi_1)} = \frac{f^{(2)}(\xi_2)}{g^{(2)}(\xi_2)}=\cdots =\frac{f^{(n+1)}(\xi_{n+1})}{g^{(n+1)}(\xi_{n+1})}\,.$$

And now we can set $\xi_b=\xi_{n+1}.$

Essentially the same argument works on $[a,x_0]$ to give us $\xi_a$ satisfying $(3)$.

This proves the Hybrid Lemma in the case where $g^{(n+1)}$ has no zeros in $(a,b)\setminus\{x_0\}.$

Otherwise, we have to go back to $(1)$ and $(2)$ instead. Does anything change?

Let's look next at the case where $g(b)=0.$ Then the RHS of $(2)$ is $0$. But we can apply Rolle's Theorem repeatedly to $g$ and its derivatives to show that there are $\xi_k$ $(k=1,\dots,n+1)$ with $b>\xi_1>\xi_2>\cdots>\xi_{n+1}>x_0$ such that $g^{(k)}(\xi_k)=0\,.$ We can take $\xi_b=\xi_{n+1}$, and equality holds in $(2)$ with both sides being $0$.

Thus we can assume that $g(b)\neq 0.$

By symmetry in $f$ and $g$, we can also assume that $f(b)\neq 0$

Applying CMVT on $[x_0,b]$ gives us a $\xi_1 \in (x_0,b)$ with

$$f(b) g'(\xi_1) = g(b) f'(\xi_1)\,.$$

Then there is $\xi_2 \in (x_0,\xi_1)$ with $f'(\xi_1)g''(\xi_2)=g'(\xi_1)f''(\xi_2)$.

However, if $f'(\xi_1)=g'(\xi_1)=0$, this tells us nothing at all about $f''(\xi_2)$ and $g''(\xi_2).$

So that approach doesn't work!

So let's just go back to $(2)$ and use Rolle's Theorem repeatedly instead (as in the proof of the CMVT and in some proofs of Taylor's Theorem), starting with a suitable linear combination of $f$ and $g$.

Define $h:[a,b] \to \R$ by

$$h(t)= f(b)g(t)-g(b)f(t)\,.$$

Then $h$ is continuous on $[a,b]$ and $h$ is $n+1$ times differentiable on $(a,b)$ with

$(*)\Jdisplay h^{(k)}(t)= f(b)g^{(k)}(t)-g(b)f^{(k)}(t)\,$

for all $t \in (x_0,b)$ and $k\in\{0,1,2,\dots,n+1\}.$

Our original assumptions on $f$ and $g$ give us that $h^{(k)}(x_0)=0$ for $k=0,1,\dots,n.$

We also have $h(b)= f(b)g(b)-g(b)f(b) = 0.$

(This is why we chose these particular coefficients for $f$ and $g$ in the first place!)

Looking back to the argument we gave above in the case when $g(b)=0,$, but working with $h$ instead, we see that we can find $\xi_k$ $(k=1,\dots,n+1)$ with $b>\xi_1>\xi_2>\cdots>\xi_{n+1}>x_0$ such that $h^{(k)}(\xi_k)=0\,.$

Set $\xi_b=\xi_{n+1},$ so that $h^{(n+1)}(\xi_b)=0$. By $(*)$, we have

$$f(b)g^{(n+1)}(\xi_b)-g(b)f^{(n+1)}(\xi_{b})=0\,,$$

and so $(2)$ follows.

We obtain $\xi_a$ satisfying $(1)$ in the same way by working with $f(a)g(t)-g(a)f(t)$ on $[a,x_0]$.

As a corollary, we now obtain a Hybrid Taylor Theorem (which can also be described as a two-function Taylor Theorem)

Corollary (Hybrid Taylor Theorem)

Suppose that, in addition to our initial assumptions on $f$ and $g$, that $n\in\{0,1,2,\dots\},$ and that $f$ and $g$ are $n+1$ times differentiable on $(a,b).$

Define $R_f$ and $R_g$ to be the $n$th remainder functions for $f$ and $g$ when looking at the Taylor polynomials for $f$ and $g$ centred at $x_0.$ So, for all $t \in [a,b]$, we have

$$R_f(t)=f(t)-\sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!} (t-x_0)^k$$

and

$$R_g(t)=g(t)-\sum_{k=0}^n \frac{g^{(k)}(x_0)}{k!} (t-x_0)^k\,.$$

Then there exist $\xi_a \in (a,x_0)$ and $\xi_b \in (x_0,b)$ such that

$(1')\Jdisplay R_f(a) g^{(n+1)}(\xi_a) = R_g(a) f^{(n+1)}(\xi_a)$

and

$(2')\Jdisplay R_f(b) g^{(n+1)}(\xi_b) = R_g(b) f^{(n+1)}(\xi_b)\,.$

If $g^{(n+1)}(t)\neq 0$ for all $t \in (a,b) \setminus \{x_0\}$ and (hence) $R_g(a)\neq 0$ and $R_g(b)\neq 0$, then we have

$(3')\Jdisplay \frac{R_f(a)}{R_g(a)}= \frac{f^{(n+1)}(\xi_a)}{g^{(n+1)}(\xi_a)}$

and

$(4')\Jdisplay \frac{R_f(b)}{R_g(b)}= \frac{f^{(n+1)}(\xi_b)}{g^{(n+1)}(\xi_b)}\,.$

Proof

This is immediate from the Hybrid Lemma, because $R_f$ and $R_g$ satisfy the conditions of that lemma, while $R_f^{(n+1)} = f^{(n+1)}$ and $R_g^{(n+1)}=g^{(n+1)}.\qquad\square$

Note

The usual version of Taylor's Theorem (with Lagrange's form of the remainder) can be obtained immediately from this Hybrid Taylor Theorem by taking $g(t)=(t-x_0)^{n+1}$, in which case $R_g=g$ and $R_g^{(n+1)}(t)=(n+1)!$ for all $t$. Then $(1')$ gives

$$(n+1)! R_f(a) = (a-x_0)^{n+1} f^{(n+1)}(\xi_a)\,,$$

and $(2')$ gives

$$(n+1)! R_f(b) = (b-x_0)^{n+1} f^{(n+1)}(\xi_b)\,.$$

For all $x \in [a,b]$ with $x>x_0$, we can apply this result on $[a,x]$ instead to see that there is a $\xi$ strictly between $x_0$ and $x$ such that

$$R_f(x)= \frac {f^{(n+1)}(\xi)}{(n+1)!} (x-x_0)^{n+1}\,.$$

The same applies if $x \in [a,b]$ with $x<x_0$, working on $[x,b]$ instead.

An alternative is to work with one-sided derivatives, and then you can allow $x_0$ to be at an end-point of the interval.