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RolyPoly

  • 2 years ago

How to determine the differentiability? For example, if the function is not continuous at \(x=x_0\), it's not differentiable at \(x=x_0\) And if the function is not ''smooth'' at a point, it's not differentiable at that point. Any other conditions?

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  1. RolyPoly
    • 2 years ago
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    FYI, (A) is not ''smooth'' at certain points while (B) is ''smooth'' |dw:1347030705975:dw|

  2. RolyPoly
    • 2 years ago
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    |dw:1347030827486:dw|

  3. experimentX
    • 2 years ago
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    |dw:1347031408321:dw|

  4. RolyPoly
    • 2 years ago
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    Yup, but apart from the above two cases, is there any condition that we can know the function is not differentiable (at certain points) ?

  5. mathmate
    • 2 years ago
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    |dw:1347072880969:dw|

  6. RolyPoly
    • 2 years ago
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    Aren't the above for cases the ones I've mentioned? For the top two cases, the limit at a certain doesn't exist. For the bottom two cases, there is a sharp point (not smooth at those sharp point) Are there any other cases apart from these two conditions?

  7. Jemurray3
    • 2 years ago
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    As long as the function is continuous at the point in question and the slope of the tangent line to the curve approaches the same value when you approach it from both the positive and negative direction, it's differentiable there.

  8. RolyPoly
    • 2 years ago
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    Thanks @Jemurray3 ! I discovered that I asked a silly question!

  9. mathmate
    • 2 years ago
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    @RolyPoly Curiously in the bottom left case, the limit exists (shape is called a cusp), they are both infinite from the left and from the right. In addition, the point is defined (y is finite). Obviously the function is not differentiable even though it is continuous and the limit exists. I have included some case in which the function is discontinuous or where the limit does not exist for the pure purpose of see the different forms undifferentiable functions can take. It is good that you recognize rapidly the reason of undiffentiability.

  10. RolyPoly
    • 2 years ago
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    @mathmate Would you mind explaining more on the bottom left case? I think it's not differentiable as I see a sharp point there. But I'm not quite sure if the limit at the point really exists. Do you minding helping a bit?

  11. Jemurray3
    • 2 years ago
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    If you're talking about the limit of the slope of the function, then by definition it does not exist if it tends toward infinity... and besides, from the left it approaches negative infinity and from the right it approaches positive infinity.

  12. RolyPoly
    • 2 years ago
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    |dw:1347083821366:dw| Is that direction right? or wrong?

  13. Jemurray3
    • 2 years ago
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    Oops I meant the other way around... from the left the limit of the slope goes to infinity and from the right it goes to negative infinity. Unless it's a corner rather than a cusp.... either way, it's not differentiable. As a general rule, if you can't make the function look like a straight line by zooming in farther and farther, it can't be differentiated at that point. Which, by the way, happens to be the condition for the curve to be a 1-dimensional manifold.... but don't worry about that now :)

  14. RolyPoly
    • 2 years ago
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    The problem is that.. how to find the limit at that point? :( |dw:1347084251150:dw|

  15. Jemurray3
    • 2 years ago
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    Of the slope or of the function? It would depend on what the explicit representation of the function is..... like x^{-1/3} for instance

  16. Jemurray3
    • 2 years ago
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    Is discontinuous, as is its derivative.

  17. RolyPoly
    • 2 years ago
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    at x=0? or generally speaking?

  18. Jemurray3
    • 2 years ago
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    just at the origin

  19. RolyPoly
    • 2 years ago
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    Okay, that's because when x=0, f(0) is undefined, limit at that point doesn't exist, hence its derivative, right?

  20. Jemurray3
    • 2 years ago
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    Yep.

  21. RolyPoly
    • 2 years ago
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    Thanks! But @mathmate mentioned that limit of the bottom left one exists. However, it's not differentiable. I don't understand this point actually. (Well, I do understand when a graph of this is provided as a see a sharp point there, but I don't know how to figure the limit out and draw that conclusion). Can you, or anyone, explain that example?

  22. Jemurray3
    • 2 years ago
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    It depends on what function you were modeling in the bottom left. What was it?

  23. RolyPoly
    • 2 years ago
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    I'm sorry but I don't know. It's just an example provided by mathmate which I'm having trouble with.

  24. Jemurray3
    • 2 years ago
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    Depending on the function the limit of the slopes may exist or they may tend toward infinity (either positive or negative) but either way they will not be the same if you approach from the left and the right separately. Therefore, the derivative is not well-defined at that point. Obviously the derivative is not well-defined if the function itself is discontinuous as well.

  25. RolyPoly
    • 2 years ago
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    So, in this case, the bottom left example is not differentiable as the slope of the function tends to infinity when x approaches that point ( from either the left side or the right sides). Right?

  26. mathmate
    • 2 years ago
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    @rolypoly @jemurray3 Here are some further information for the lower-left example (cusp). An example would be \( \large f(x)=10-x^\frac{2}{3} \) Continuity, which is necessary for differentiability, is satisfied here, since f(0) \( \in R \) and is finite, and f'(0+)=f'(0-). However f(x) is not differentiable because Darboux's theorem requires that, for f(x) to be differentiable between a and b, f'(x) must take on every value between f'(a) and f'(b). If we set a=\( -\epsilon \) and b=\( +\epsilon \), where \( \epsilon \) is an arbitrarily small number \( \ne 0\), we see immediately that Darboux's theorem is not satisfied, and therefore f(x) is not differentiable at x=0. This is a case that illustrates that continuity is a necessary but not sufficient condition for differentiability.

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  27. mathmate
    • 2 years ago
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    Errata: for continuity, meant: \( lim_{x \rightarrow \ 0-} f'(x) \) = \( lim_{x \rightarrow \ 0+} f'(x) \) but hopefully should have been understood from the context.

  28. Jemurray3
    • 2 years ago
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    Where you lose me is the part where \[ \lim_{x \rightarrow 0-} f'(x) = \lim_{x\rightarrow 0+} f'(x)\] since neither of those limits converge to a finite number.

  29. mathmate
    • 2 years ago
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    Continuity requires only that lim f(x-) = lim f(x+), which is satisfied by both y=|x| and y=10-x^(2/3). For y=|x|, \( lim f(x-) \ne lim f(x+) \), which makes it clearly not differentiable. For \( \large y=10-x^{\frac{2}{3}} \), it passes the test, but does \( \textit not \) satisfy Darboux's theorem. This is the point how important it is to know Darboux's theorem for differentiability.

  30. Jemurray3
    • 2 years ago
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    And yet again I'm forced to ask what makes you say that y = 10 - x^(2/3) passes such a test if *neither of those two limits exist*

  31. mathmate
    • 2 years ago
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    The function passes the test for continuity (just like y=|x| did), so it is continuous. It also passes the test for f'(x-)=f'(x+), which means that the limit of f'(x) exists, although it equals infinity. The function passes \( \textit both \) tests, but it does not satisfy Darfoux's theorem, so it remains not differentiable.

  32. Jemurray3
    • 2 years ago
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    The if the limit tends toward infinity, the limit does not exist. Period. "Infinity" is not a quantity in the real number system.

  33. Jemurray3
    • 2 years ago
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    The definition of a limit implies a finite quantity.

  34. mathmate
    • 2 years ago
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    The limit of f'(x) can be infinite and still defined, as long as it is the same value approached from both sides.

  35. Jemurray3
    • 2 years ago
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    I hate to sound like a broken record, but that's just not true. Go look up the mathematical definition of a limit. It would look something like this: \[\text{If } \exists \delta>0 \text{ such that, given some } \epsilon >0, |x-x_0|<\delta \text{ implies that }|f(x)-L|<\epsilon\] \[\text{ where } \delta,\epsilon,L \in \mathcal{R}, \text{ then L is called the limit of f as x approaches }x_0. \]

  36. mathmate
    • 2 years ago
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    You are perfectly correct in saying that the limit does not exist at x=0, mathematically speaking, i.e. according to the definition. Your definition is valid and correct. There are also other definitions for the special cases for infinity and negative infinity as limits, such as: "We say that f(x) approaches infinity as x approaches \( x_0 \), and write \[ \lim_{x->x_0}{} f(x)=\infty \] if for every positive real number B there exists a corresponding \( \delta > 0\) such that for all x \( 0<|x-x_0|<\delta => f(x) > B \) " I realize that the definition does not justify the existence of the limit, but nevertheless explains my loose language. My apologies.

  37. Jemurray3
    • 2 years ago
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    That's fine, and I don't mean to resort to nitpicking terminology but when I said that f'(0+) = f'(0-) was, assuming continuity, sufficient to ensure differentiability, I tacitly assumed due to the definition of limit that those two quantities were finite. Would you disagree that this was sufficient? I.e. can you supply a case in which these two requirements are met but the system is still not differentiable?

  38. mathmate
    • 2 years ago
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    The point of this example is to show exactly that f'(0+)=f'(0-) is NOT sufficient for differentiability. Apart from continuity, which is shared by many of the example cases, we need to satisfy Darboux's theorem, which says that (from memory) If a real function f(x) is differentiable on the interval [a,b], its derivatives assumes all intermediate values. Basically it requires that the derivative to be continuous on the given interval. Our example does not satisfy the theorem of assuming all intermediate values. We see that f'(x-\( \delta \)) >0 and f'(x+\( \delta \)) < 0, but f'(x) is never zero on this interval. Therefore Darboux's theorem is not satisfied, and f(x) is not differentiable at x=0.

  39. Jemurray3
    • 2 years ago
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    I understand Darboux theorem is not satisfied, but it is equivalent to saying that the first derivative is continuous. But that is exactly the same as saying that f'(0+) = f'(0-)!

  40. Jemurray3
    • 2 years ago
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    It is fairly clear that neither one of us is going to convince the other. Speak to a math professor or something and if your assertion is verified then you can rest easy knowing that my understanding of the issue is flawed.

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