Second Derivative of Abs(x) | Richard Seeking Answer

[Ebolamonk3y]

:) Says who that it is not differentiable!

 

[matt grime]

Gokul, since |x| isn't differentiable everywhere, it doesn't make sense to talk of the derivative of something that does not exist, especially since the step funcntion is of the form, 0 for x<k, 1 for x=>k, for some k (right continuous, there is of course a left continuous variation, and it's use is in a formal analogy to the FTC in the integral sense, not differential); it is defined at k, which isn't like the 'derivative' of |x| whihc isn't defined at zero.

 

[matt grime]

O.K, I understand from the point of view of dealing with simple polynomials of finite order but what about if we had something like:

y = \left| \sin \left( \frac{1}{x} \right) \right|

And we wanted to know y', surely it would be plain painful dealing with it a section at a time.

not in the slightest: it is clearly not differentiable at the points where 1/x is an integer multiple of pi and the rest is obvious, depending on whether x is in an interval ((2m\pi)^{-1},((2m+1)\pi)^{-1}) or (((2m+1)\pi)^{-1},(2m\pi)^{-1}) for some integer m

you might also find it useful to write sign(x) rather than x/|x|, or |x|/x, cos that's all that's going on.

 

[Zurtex]

not in the slightest: it is clearly not differentiable at the points where 1/x is an integer multiple of pi and the rest is obvious, depending on whether x is in an interval ((2m\pi)^{-1},((2m+1)\pi)^{-1}) or (((2m+1)\pi)^{-1},(2m\pi)^{-1}) for some integer m

you might also find it useful to write sign(x) rather than x/|x|, or |x|/x, cos that's all that's going on.

lol, simple as eh? I'm obviously missing something because I would find it difficult to work out where a number is in that interval, particularly for very small values of x.

If I was to use sign(x) it would definitely not work for where f(x) = 0 and f'(x) = 0.

 

[matt grime]

floor(1/xpi) would seem useful if you were actually to want to do anything as dull as work something out explicitly

 

[Parth Dave]

|x| is not differentiable at x = 0. If you look at the definition of the derivative you can see why. The derivative is defined as the instantaneous rate of change at a given point (x). In order words, the derivative is the slope of the tangent line at that point. In |x|, at x = 0, you can create a infinite number of tangent lines. Because of that you can't find a single value for the slope of the tangent line. That is why is it non-differentiable at that point. If you wanted to find the derivative of |x| than you would have to look at it as a piece-wise function (ie, break it up into two functions). Find the derivative before x=0 and after x=0. But remember the derivative will not be defined at x=0. You can use this same method for finding the derivative of any absolute value function. Break up the function into a piece-wise function, find the domain of each of those functions, and than find the derivative in that domain.

 

[Ebolamonk3y]

Well... I guess Finney can be wrong... In his Calc book he explicity stated that |x| derivate is what someone already said on here...

 

[Zurtex]

floor(1/xpi) would seem useful if you were actually to want to do anything as dull as work something out explicitly

Riiight, well I seem to have failed in making my point all together. My point was that dealing with it a section at a time for a sufficiently hard equation would be very difficult, so here is a sufficiently ridiculous example:

y = \left| \sin \left( \frac{\sqrt{2}}{x^{ \frac{\pi}{e}}} \right) + \frac{e^{x^\frac{1}{2}}}{\pi} \right| - |x|

 

[matt grime]

finney is about as trustworthy as nixon (well perhaps not, but it certainly isn't very good), but perhaps it states what the derivative is at all points except 0.

and, zurtex, can you tell me what the derivative is of your function with your formula? since that requires you to know where the zeroes are, just as my notional method would do, i don't see that there's any difference at all, in fact. note, that as the function you've given is not of the form |f|, then your formual does not apply, unless you prove some further results about additivity, which don't seem obvious.

 

[Zurtex]

Fine fine, but do you at least except that my method works?

 

[matt grime]

i believe I've repeatedly said that except at points where things are zero your method gives the right answer, but that at zeroes you still need to be careful, and not the additional comment that the example you give is not of the form |f| so it doesn't constitute a validd function for your method

 

[Zurtex]

i believe I've repeatedly said that except at points where things are zero your method gives the right answer, but that at zeroes you still need to be careful, and not the additional comment that the example you give is not of the form |f| so it doesn't constitute a validd function for your method

kk thanks I might look at writing something up as a method to actually make it easy.

But you can simply apply the standard result:

\frac{d}{dx}(u-v) = \frac{d}{dx}(u) - \frac{d}{dx}(v)

And then use the formulae I gave earlier to work out both.

 

[matt grime]

how do you know that formula applies at points where u and v aren't differentiable? hint: you don't, it doesn't

 

[Zurtex]

how do you know that formula applies at points where u and v aren't differentiable? hint: you don't, it doesn't

What?

Are you saying that:

\frac{d}{dx} \left( \frac{1}{x} - x^2 \right) \neq \frac{-1}{x^2} - 2x

Because it is not differentiable at the point x = 0

 

[matt grime]

the function you've written down isn't even defined at 0, so i think asking for its derivative there is a little unreasonable.

i am not talking about the points where a function's derivative is defined, but the points where it is not defined.

 

[Zurtex]

the function you've written down isn't even defined at 0, so i think asking for its derivative there is a little unreasonable.

i am not talking about the points where a function's derivative is defined, but the points where it is not defined.

Sorry I clearly don't have enough understanding of calculus because I don't see what is wrong, I'll let it go now.

 

[MiGUi]

f(x) = |x|

lim f(x) = 1
x -> 0+

lim f(x) = -1
x -> 0-

Lateral limits are not equal, so the function is not derivable in x = 0.

 

[CodeAvenger]

the second derivative of |x| at x=0 is infinite.
and the first derivative at x=0 is undefined

 

[Maddox731]

I'd like to start by saying I agree with everything said above so far, |X| is not differentiable at X=0, but from the OP he's interested in a Quantum Mechanical situation. In QM we can (nay, we are compelled to based on our need for a rigged hilbert space as a basis) use generalized distributions and not functions.

Graphically picture |X|, notice that, as above, its derivative at X<0 is -1 and X>0 is +1. Granted. What about X=0? Well, its not differentiable there. As you zero in on it (As someone has also done above) you notice X->0+ = 1 and X->0- = -1. What generalized distribution do you know looks like this? The Heavyside function! θ(X) ! (There are vastly more rigorous solutions out there using these).

Now what is the derivative of the Heavyside function? Naturally it is also discontinuous (these distributions sure are troublesome), but if you were to examine its slope at 0 it jumps from -1 to 1 rather quickly. This suggests a function who is 0 at all points |X|>0 except exactly at 0 where it diverges. This looks like another friendly distribution, the Dirac delta function!

I apologize for my lack of rigor above. :-)

Edit: Wow sorry to bump this up from 2004, it was just a result on the ol' google search, figured I would add what I learned in QM.