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 proof 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.