Second-order MVT

Firstly, what is a second-order mean value theorem? Does it refer to things of the form

f(a + h) = f(a) + h f(a) + R_1(h)

? In particular, for the question below, which form of the remainder will be most helpful?

(Q11 on this sheet)

Let

f : [-1, 1] \to \mathbb{R}

be continuous and twice-differentiable on the (open, closed, resp.) interval. Let

\displaystyle \phi(x) = \frac{f(x) - f(0)}{x}

, with

\phi(0) = f'(0)

. Using a second-order mean value theorem for

f

, show that

\displaystyle \phi'(x) = \frac{1}{2} f''(\theta x)

for some

0 < \theta < 1

. (etc.)

Using two different forms of the remainder, I can get two erroneous proofs:

First, using the Lagrange remainder,

\displaystyle R_1(x) = \frac{1}{2}x^2 f''(\theta x)

, we get

\displaystyle \frac{1}{2} f''(\theta x) = \frac{1}{x^2}( f(x) - f(0) - x f'(0) )

. So, differentiating

\phi(x)

, for

x \ne 0

, we get:

Unparseable latex formula:

\displaystyle \begin{aligned}[br]\phi'(x) & = \frac{f'(x)}{x} - \frac{f(x) - f(0)}{x^2} \\[br]& = \frac{f'(x) - f'(0)}{x} - \frac{f(x) - f(0) - x f'(0)}{x^2} \\[br]& = \frac{f'(x) - f'(0)}{x} - \frac{1}{2} f''(\theta x)[br]\end{aligned}

Now, at this stage, one would be tempted to rewrite

\displaystyle \frac{f'(x) - f'(0)}{x} = f''(\theta x)

, but I have no guarantee that this θ is the same as the one above. But if I assume that is so, I get the result I want.

This second "proof" is more interesting: Using

\displaystyle R_1(x) = \frac{x^2}{2} \frac{f'(\theta x) - f'(0)}{\theta x}

, we can get

\displaystyle \phi(x) = \frac{f'(\theta x) - f'(0)}{2\theta} + f'(0)

. At this stage, my friend differentiated this expression to get

\displaystyle \phi'(x) = \frac{1}{2} f''(\theta x)

. I then pointed out that

\theta

is actually a function of

x

, so he would need to apply the chain rule etc. He then pointed out that we don't even know if

\theta

is continuous, let alone differentiable. He then assumed it was differentiable and applied the chain rule and got a few extra terms, which had to sum to zero for the result to be correct. But, according to him, the only way that could happen is if

\theta'(x) = 0

. But the result is for arbitrary

x

, so this would imply

\theta

is constant. But this means there's one constant

\theta

that works for any

x \in [-1,1]

, which is simply too good to be true. Another friend tried some different assumptions and obtained a differential equation for

\theta

, and solving it gave something like

\displaystyle \theta(x) = \frac{1}{2} + \frac{c}{x}

, which is either patent nonsense, or too good to be true again.

I think I'm missing something here. :-/

Reply 1

Dystopia

10

I've only looked at your first one, but you need to think more carefully about what you're doing. It might be helpful to work back from the answer to see where it comes from.

Reply 2

Zhen Lin

OP

12

Well, that's one of the problems: I have absolutely no idea where it comes from.

I've just noticed that I can get

\displaystyle \frac{\phi(x) - \phi(0)}{x} = \frac{R_1(x)}{x^2}

out. Using the Lagrange remainder, and doing a MVT substitution, I get

\displaystyle \phi'(\chi) = \frac{1}{2} f''(\xi)

, and both

\chi / x , \xi / x \in (0, 1)

, but how do I know that

0 < \theta = \xi / \chi < 1

?

Reply 3

ukgea

6

I haven't looked into your proofs too much, but I have to say I feel you're making things more complicated than they are.

Hint: Differentiate

\phi(x)

and thus write out exactly what you need to show (in a Taylor's theorem-like form), then rearrange the terms until it becomes clear.

Reply 4

Zhen Lin

OP

12

Hmmm. I sort-of tried that already, it doesn't come out, at least not immediately. (See first "proof".)

Either I'm worrying about things I don't have to worry about, or I'm missing something entirely.

Reply 5

ukgea

6

Zhen Lin

Hmmm. I sort-of tried that already, it doesn't come out, at least not immediately. (See first "proof".)

Either I'm worrying about things I don't have to worry about, or I'm missing something entirely.

Hmm. I think this is just something you would have got eventually anyway, so let me put it this way: What you want to show is

\displaystyle \frac{x^2}{2}f''(\theta x) = x f'(x) - f(x) + f(0)

agreed? There are basically two ways you could rearrange this to make it remotely useful, one of those ways work.

Reply 6

Zhen Lin

OP

12

Aha, I think I see it. What a twisted thing to do... I guess this is what you meant?

Spoiler

Thanks!

Reply 7

DFranklin

18

For what it's worth, I think this is one of those things where once you get your mind running down the "wrong path", it is very hard to see the right one. I saw your post and couldn't see how to progress at all. And although I am not the analyst I used to be, it's still pretty unusual for me to get stuck on this level of question.

And then I looked again today, not looking at your working and I found it came out quite easily.

Reply 8

Zhen Lin

OP

12

Yes, it seems to be. I think, it's because while heading down the wrong path, one sees things that are so close to the result desired, to the point where one thinks that one just needs one more lemma to get it down... certainly that's what happened to me for this question.

Second-order MVT

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