Question

Representing higher order derivatives in terms of first derivatives problems:

Answer 1

\[\Large f'(x) = \lim_{h \to 0} \frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}\] This is my derivative definition, so if we apply it again we get the second derivative like this: \[\small f''(x) = \lim_{h \to 0} \frac{1}{h}(\frac{f(x+\frac{h}{2}+\frac{h}{2})-f(x+\frac{h}{2}-\frac{h}{2})}{h}-\frac{f(x-\frac{h}{2}+\frac{h}{2})-f(x-\frac{h}{2}-\frac{h}{2})}{h})\] simplify a little we get: \[f''(x) = \lim_{h \to 0} \frac{1}{h^2}[f(x+h)+f(x-h)-2f(x)]\] So with a little rearranging we can show an interesting fact that the concavity is proportional the the difference between the function at a point and the average of the two adjacent points, but that's not really what I'm looking for, I want to take the third derivative now and get a result, so we continue on and we can simplify to something that looks like this:

\[\ f'''(x) = \lim_{h \to 0} \frac{1}{h^2}\frac{f(x+\frac{3h}{2})-f(x-\frac{3h}{2})}{h}-\frac{3}{h^2}\frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}\]

Answer 2

Can we write this final thing in terms of just ordinary derivatives? It is tempting to just rewrite it by a change of variables, but I don't know if it's possible or how I would do it.

Answer 3

If we apply the derivative a second time, I think we should end up with two limits.

Answer 4

They may be equivalent, but I'm not completely sure just yet.

Answer 5

So instead of combining the h's use separate variables, h, then k, etc? I am not sure either but it is an idea to play with.

Answer 6

Start like this, then use definition of derivatives:\[
\lim_{k\to 0}\frac{f'(x+k)-f'(x)}{k}
\]

Answer 7

By changing variables and separating the limit into two parts we can have:
\[\ f'''(x) = 216\lim_{h \to 0} \frac{1}{h^2}\frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}-3\lim_{h \to 0}\frac{1}{h^2}\frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}\]

however I don't even know if I'm technically allowed to separate this limit or not.

Answer 8

I think that will just give essentially the same thing @wio I just kind of like looking from this perspective since I can sort of see symmetry a little easier.

Answer 9

Well for one, does the order of taking limits matter? You're assuming it doesn't when you merge them.

Answer 10

Order of taking limits? Is it even possible to take the second derivative before the first derivative?

Answer 11

I mean letting \(k\) goto \(0\) before \(h\) dies.

Answer 12

Well what is the difference between k and h?

Answer 13

If I switched them when you weren't looking you wouldn't tell. It's like if I have x and want to add 2. I can either add 1 first and then add 1 or I can add 1 then add 1 the other way around, and no one can tell the difference I think.

Answer 14

Start like this, then use definition of derivatives:\[
\lim_{k\to 0}\frac{f'(x+k)-f'(x)}{k} =\lim_{k\to 0}\lim_{h\to 0}\frac{f(x+h+k)-f(x+k)-f(x+h)+f(x)}{hk}
\]

Answer 15

Sure, now replace h and k with each other, they are completely symmetric: \[
\lim_{k\to 0}\lim_{h\to 0}\frac{f(x+h+k)-f(x+k)-f(x+h)+f(x)}{hk} = \
\
\]\[ \lim_{h\to 0}\lim_{k\to 0}\frac{f(x+k+h)-f(x+h)-f(x+k)+f(x)}{kh}
\]

Answer 16

So if they're symmetric then if follows that we can let \(h=k\) I suppose?

Answer 17

I think so, since I can imagine bringing h to 0 is the same as making k to 0 so in the hk-plane we should be able to just go straight to 0 in a single line

Answer 18

I have seen it done before by a differential equations book I owned, but then again just because it works doesn't necessarily mean anything.

Answer 19

Were aren't taking them at the same time, we're taking them in order, which does make things easier.

Answer 20

We've already assumed that \(f'(x)\) exists.

Answer 21

Yeah this is all done based on the idea that all the higher derivatives we need exist, that's not important to worry about.

Answer 22

I think since we assume the derivatives exist, then we can use: \[
\lim_{x\to a}g(x) = b\land \lim_{x\to b}f(x) = L \implies \lim_{x\to a}f(g(x)) = L
\]

Answer 23

In the case the function behaves relatively nicely, the two-limits of a second order derivative can be reduced to a single limit--but keep in mind this is the exception, not the rule.
$$f''(x)=\frac{df'}{dx}=\lim_{\delta x\to0}\frac{f'(x+\delta x)-f'(x)}{\delta x}$$
... is the general definition of the second derivative of a differentiable function \(f:\mathbb{R}\to\mathbb{R}\).

If you're interested in similar limits for higher order derivatives then you should check out finite differences.

http://en.wikipedia.org/wiki/Finite_difference

Answer 24

http://en.wikipedia.org/wiki/Finite_difference#Higher-order_differences

In the limit as the step size or spacing goes to 0, the general formulae for the higher-order finite differences agree with the corresponding derivatives (for sufficiently well-behaved functions).

Answer 25

Okay, I'm curious as to what the exceptions are.

Answer 26

well behaved means \(C^1\)?

Answer 27

I think the exceptions would fall along the lines of \(f(x)\) is not differentiable and yet the higher order difference limit just so happens to exist.

Answer 28

I honestly didn't think too deeply about any particular conditions for 'well-behaved' but I figured I'd play it safe :-p I'll give it some thought.

Answer 29

but indeed, you can find more of these single-limit formulae using the n-th order finite differences and simply taking their limit as the step size goes to 0.

Answer 30

You're right. Proving anything complicated concerning limits ends but being a major battle.

Answer 31

Differentiable is a very high criterion, typically speaking.