What do you mean? If we have any continuous function, then we do not have necessarily a derivative. So how could we make a statement about derivatives?feynman1 said:Darboux theorem says a derivative function must have an intermediate valuetheoremwithout requiring the derivative function to be continuous. Why is this property not true for any continuous function in its intermediate value theorem?
Not at all. You asked:feynman1 said:This is to compare intermediate value thm and Darboux thm. Clear?
[Darboux: ##f'## takes every value between ##f'(a)## and ##f'(b)##]feynman1 said:Why is this property ...
Continuous functions in general do not have derivatives. Hence the property cannot hold since the statement is void.feynman1 said:... not true for any continuous function in its intermediate value theorem?
The reason is, that ##f'## doesn't jump! ##f## is differentiable, so it may have kinks, but it does not have gaps. It goes roughly along the following lines:feynman1 said:No. Why does a derivative function possesses a property of taking any intermediate value without declaring the derivative is continuous?
why can f' not have any gaps while f' can have kinks?fresh_42 said:##f## is differentiable, so it may have kinks, but it does not have gaps.
This is a consequence, not a cause. However, it illustrates why it works. It's a heuristic.feynman1 said:why can f' not have any gaps while f' can have kinks?
is there any counter example?Office_Shredder said:I think your takeaway should be that the derivative is *basically* continuous. Do you know an example where it is not? I think it's pretty instructive to just meditate on one for a while.
yes, is there any intuition to understand this consequencefresh_42 said:This is a consequence, not a cause. However, it illustrates why it works. It's a heuristic.
You take a value in the gap and construct a contradiction to Darboux. It is a bit messy with all the theorems that basically look like one another.
feynman1 said:is there any counter example?
this example is to show that its derivative jumps or not?Office_Shredder said:##x^2\sin(1/x)##
Basically the idea is that it is bounded by a parabola, so the derivative at ##x=0## is 0, but it wiggles around really fast at you get near 0, so you can get not close to 0 derivatives no matter how close to 0 you get.
feynman1 said:this example is to show that its derivative jumps or not?
Do you mean although the derivative is discontinuous, the derivative is always able to take any intermediate values demanded?Office_Shredder said:No, the derivative does not jump, but it is discontinuous at x=0. You should try graphing this function and its derivative and thinking for a while about what is going on.
feynman1 said:Do you mean although the derivative is discontinuous, the derivative is always able to take any intermediate values demanded?
For a discontinuous function, the function won't in general be able to take any intermediate values demanded unless specially designed.
yes, the derivative is discontinuous at x=0Office_Shredder said:Have you looked at the derivative of this function near 0? Draw a graph, and evaluate the derivative at 0 yourself.
feynman1 said:yes, the derivative is discontinuous at x=0
why is this feature not in continuous functions?Office_Shredder said:So this is what the darboux theorem is saying. The only way a derivative can be discontinuous is it can wiggle around and take a bunch of different values near x=a, but still be defined at x=a
If you "differentiate" the continuous function ##x \longmapsto |x|## then you get a gap of ##+2## at zero. But if we start with a differentiable function ##f##, say at ##0##, then we havefeynman1 said:why is this feature not in continuous functions?
feynman1 said:why is this feature not in continuous functions?
but when talking about discontinuity in a derivative, it is about any gap in c rather than ch? so in your example if c is different from c' the derivative is discontinuous?fresh_42 said:If you "differentiate" the continuous function ##x \longmapsto |x|## then you get a gap of ##+2## at zero. But if we start with a differentiable function ##f##, say at ##0##, then we have
$$
f(h)=f(0) + \varphi (h) + r(h) = f(0)+c\cdot h +r(h)
$$
with a small remainder term ##r(h)## and a linear function ##\varphi \, : \,x\longmapsto c\cdot x .## Now ##c\cdot h## is still small for small ##h## and our function value ##f(h)## is still close to ##f(0)##. I.e. there is no way to get a gap of ##+2## for small and smaller ##h.## We can have as derivative ##\varphi (x)=cx## from the left and ##\varphi' (x)=c'x## from the right, which is discontinuous if ##c\neq c'.## This means we have a discontinuity of the derivative. However, ##ch## and ##c'h## are close for small ##h##.
Yes. My example has (excuse the sloppy notation): ##f'(0^+)=c## and ##f'(0^-)=c'##. So we have a gap of ##|c'-c|.## But the function values ##f(h)## and ##f(-h)## are close to ##f(0)## because the derivative is multiplied by ##\pm h.## This means that there is still some control on ##f(x)## by the simple fact that it is differentiable. A property we do not have for only continuous functions.feynman1 said:but when talking about discontinuity in a derivative, it is about any gap in c rather than ch? so in your example if c is different from c' the derivative is discontinuous?
You can take a look at the proof of the Darboux Theorem. One proof that you can find on Wikipedia uses two theorems: A continuous function on a closed interval achieves a maximum and a minimum and the derivative of a differentiable function at a local extremum is zero.feynman1 said:Thank you, but this darboux thing says any intermediate derivative can be taken, not its function value. How to understand that?
feynman1 said:Thank you, but this darboux thing says any intermediate derivative can be taken, not its function value. How to understand that?