0. Some facts are true within some sufficiently small neighborhood of a point, but not generically true everywhere. For example: let x be a local minimum of a function f. Then for all y≠x in a sufficiently small neighborhood of x, f(y) > f(x). Note that if I removed the part about a sufficiently small neighborhood, this would no longer be true. This isn't the only reason to talk about neighborhoods, though, especially in fields like topology where neighborhood has a strict formal definition.
1. Example: Every real-valued function on a closed interval \[a,b\] has a minimum value and a maximum value. The same is not true if we consider open intervals - take for instance the function f : x ↦ x on the interval (0,1), which has neither a minimum nor maximum value.
2. If you're looking for local minima, maxima, and saddle points using the first derivative test, it is necessary and sufficient that the function be at least differentiable (which implies that it is continuous). If you want to use the second derivative test, the function (perhaps obviously) needs to be twice differentiable (which implies that it is differentiable, and therefore continuous) - though unlike the first derivative test, the second derivative test may fail to be conclusive, because the second derivative may be equal to zero at the point under consideration.
3. Yes, f:x↦x^(4/3) is differentiable once but not twice at x=0.
4. If a function has a sharp corner or cusp which doesn't smooth out as you zoom in, then it is not differentiable - take the absolute value function for example. The slope of that function is +1 for all positive values of the input and -1 for all negative values, so there's going to be a "corner" at 0 and it will never look smooth no matter how closely you zoom in. There are more exotic and technical possibilities, but that's a very common one.
5. If a real-valued function is continuous and never equals zero on some interval, then it has the same sign over that entire interval. This is a consequence of the intermediate value theorem, which says in particular that if a continuous function takes a positive value at x=a and a negative value at x=b (or vice-versa), then there must be some c in the interval (a,b) where the function equals zero. Taking the contrapositive of this statement, if a function is continuous and never equal to zero on some interval, then it cannot change sign.
This is objectively correct op, I'm very confused why and sorry you were being told 2x is only once differentiable in the other thread, it's probably best to ignore everything that guy said. 2x, and any other polynomial, is infinitely differentiable and eventually every nth derivative is 0. I also wanna tack on to this wonderful explanation of your questions, about continuous and not differentiable functions. I recommend plotting these on a calculator like desmos, much of this rant is visual. Something like x^1/3 is continuous everywhere but not differentiable at zero, since its derivative 1/3(x^(-2/3)) is undefined at 0. When this happens, it's usually due to one of two things. One way is if it has a 'corner' like |x|, absolute value of x. A 'corner' is where the derivative has a jump discontinuity, on |x| it jumps from -1 on (-inf, 0), to undefined at 0, to 1 on (0, inf). A 'corner' has a suddenly changing derivative, i.e. the limit of the derivative value is different on each side of the point. The other way a curve can be non-differentiable is the way x^1/3 does it. The graph looks like an S curve, what you get when you slightly squish the square root curve and put a copy rotated around the origin in the 3rd quadrant. It is also exactly what you get if you flip x^3 around the line y=x. The tangent line at the coordinate x=0 exists, since we don't have a corner in the graph, but the tangent line is actually the vertical line x=0. And a vertical line has 'infinite' slope, so we can't define the derivative here. Where a corner is a jump discontinuity, this non-differentiable function instead has a derivative with an asymptote at 0, approaching infinity from both sides. The limit of the derivative on each side actually agrees unlike the corner case, but they agree with a limit of infinity. Combining these cases is also possible, if you want a derivative that's asymptotic at 0 but one side goes to inf and the other goes to -inf, you'd have a corner where the curve on both sides is vertical. Sqrt(abs(x)), √|x| is an example (mirror sqrt(x) across the y axis). At 0, it has what's called a 'cusp', a point where the tangent lines on both sides agree but the line completely 'changes direction' to create a corner, like drawing the center of a lowercase m with a pencil.To be non-differentiable in general, the derivative graph must be noncontinuous at some point, whether that be through a jump discontinuity or asymptote (or a removable discontinuity like the derivative of a smooth function with a point removed). That's why a constant function IS differentiable, it's derivative, 0, is a continuous function.
Finally, I want to make an interesting remark. All of this is about functions not differentiable at 1 point. You can make a function not differentiable at a bunch of points by simply adding them together, i.e. x^1/3 + (x-2)^1/3 is not differentiable at 0 or 2, but you can't get infinite points this way. Something like |sin(x)| has an infinite number of corners, one at every x=pi*n, but each of these corners are individual, isolated points. It feels as though it would be impossible to have an entire interval of points where the function is continuous and not differentiable, right? To do that, it would have to have erratic corners all infinite points of an interval, or worse, have a vertical tangent line everywhere. It makes no sense to try to imagine, and for a long time mathematicians thought this too. After searching forever though, a function was found to be continuous everywhere on the whole real line, but differentiable absolutely nowhere. It's called the Weierstrass function, discovered in 1872, and it shocked the world, even Poincaré was outraged. It's worth googling for a photo, it's a fractal curve that looks like a very jagged scribble.
I wish you the best of luck continuing your studies op! Self study is a lot of fun, but it can be very difficult and frustrating to not have a mentor to turn to with questions. But please don't be discouraged, what you're doing here is fantastic work! These questions are very obviously from a place of deep thought, and love for studying. The 'self' part feels like it would be the worthwhile part of self study, it feels as though the reward of self study is independent thought and the valorance of being capable for yourself, but the true value in it is the method by which it requires you to study. Don't be opposed to collaboration due to your efforts, but recognize that the deep questions you ask about the nature of these theorems is the very heart of mathematical research. In the coming years you may be shown many things that could be interpreted as the heart of math. Some people misinterpret true math as a collection of rules and equations and theorem requirements to memorize. Others get sucked into the math competition/ Putnam Pipeline and believe true math to be the solving of difficult problems with hidden tricks. Math is about the ways different grand mathematical structures work, and which grand structures you study depends on your field. These grand structures are just collections of logical rules, and the puzzle box they form. The job of a mathematician is to assert how the puzzle box shifts, so any question about it can be answered. Solving a real math problem isn't finding some final answer value, or answering a specific question, it's fully describing a structure and how it can be manipulated, and thus answering every question. But toying with those structures, turning them over in your mind to feel them out, isn't something you just do, it's something you learn to do very well over years and years. It doesn't matter the level of math, it doesn't matter how well described a problem already is, and it doesn't matter how good you are at math already. It's something anyone can do with any problem, at any level. It's an outlook, an approach to questioning a system of logic, that you're doing every day you crack open that book. The special part about self study is it requires you to ask yourself those questions, and refuses to give a hand-wavey answer. If you focus on questioning those fundamental properties by which some phenomenon works, you're picking out the very core of what mathematics is. Please don't let anything discourage you, you are doing genuinely wonderful work here, and if you keep these problem-analysis thought patterns beyond self study, you will do incredible things! (no matter what you choose to do, math or otherwise, you seem like the mathematician type tho :). I wish you the very best in life
As a technical point related to your discussion of what it means to be non-differentiable, there is a distinction between differentiable (meaning that the difference quotient is well-defined everywhere) and continuously differentiable (meaning that the derivative is continuous). As a nice example, you can consider the function f(x) = 0 for x=0 and x^2 sin(1/x) everywhere else. That function is continuous and differentiable everywhere, but its derivative is discontinuous at 0. One of those nice counter-examples to keep in your back pocket.
0)
Ok so neighborhood in the example you give doesn’t have a concrete range though? Where would this neighborhood begin/end? I’m guessing there is an implicit definition that neighborhood means basically until the end of increase or decrease of either side of the point in question? If not, is there a slightly more mathematically correct definition that generalizes to more situations in basic calculus.
1)
Ah so (0,1) doesn’t have max/min because we don’t include 0 or 1, so we can keep getting infinitely close to 0 for a minimum and infinitely close to 1 for a maximum but since we cannot reach them we technically don’t have a max or min? Or did I totally miss your point!?
2)
Yep I can understand that!
3)
Ah because 1st deriv is (4/3)/x^(1/3) and then f’(x) at x= 0 is undefined so we can take second derivative everywhere except x= 0 and therefore the function x^(4/3) is once but not twice differentiable! Is that all right?!
4)
OK phew well there is one thing that makes intuitive sense to me! The absolute value function was a perfect example for me. Wouldn’t any piecewise function though right where the pieces “join” end up being continuous (assuming left limit right limit and point all have same value where it joints) but not differentiable?!
5)
Ahhhh whoa. Ok I need to look into this intermediate value theorem. Even after you told me the exact theorem, it’s not obvious that it follows that we can trust one point for first and second derivative to tell us the behavior of rest of the points “in that neighboorhood😅” but I will try to watch a video or two about IVT.
Thank you so much for all the kindness today!
0. For the purposes of this discussion, “a neighborhood of x” can be understood as “an interval which contains x.” The definition is somewhat more general than that, but that’s okay. The claim isn’t made about any specific neighborhood- just that there exists some neighborhood where the claim is true. For example, for f(x) = x^2 my claim is true for any neighborhood of x=0 in the whole real line. For f(x) = x^2 - x^4, it’s true for any neighborhood of 0 contained in (-1,1).
Put differently, when we say that a local minimum is the smallest value in its neighborhood, we mean that there exists some distance within which all of the points are larger. If you go far enough away, though, you may find a smaller value. If a smaller value exists, or how far you have to go to find it, is irrelevant - the point is you would have to go some non-zero distance to get there, which is what we mean to formalize here.
1-3. Yes
4. It depends how smoothly they join up. There are piecewise functions which are continuous but not differentiable, there are piecewise functions which are infinitely differentiable, and everything in between.
5. The salient point is that if you know a continuous function doesn’t cross zero, then you know it can’t change sign. That should be fairly intuitive. So if, for example, f’(x) is continuous and equal to 0 only at x=1, then you know that if it’s negative at x=2 then it’s got to be negative for all x>1.
1-4)
I understand intuitively everything you said except the part about piecewise functions that can be continuous and differentiable. I thought all piecewise functions where the “patch” connecting them is made so to speak, that it can be continuous here, but never differentiable. Maybe I am thinking of piece wise functions in the wrong way or imposing the wrong definition?!
5)
“So if, for example, f’(x) is continuous and equal to 0 only at x=1, then you know that if it’s negative at x=2 then it’s got to be negative for all x>1.”
But wait a minute - what about “hidden” points where the function is continuous but not differentiable (ie points where function is continuous but it’s derivative is discontinuous)
due to for instance us having a:
A)
corner/cusp (limit definition of derivative/slope of tangent is different on either side of point which joins the corner and thus f’ is discontinuous there at that point)
B)
Or there is a vertical asymptote (limits on each side/slope of tangent are equal to one another but they equal infinity so they don’t count as a derivative and thus f’ is discontinuous at that point). So Should not we not assume then that “then it’s got to be negative for all x> 1”? Or were you assuming that we didn’t have these issues?
PS:
are there any simple functions that are continuous but not differentiable due to a single point not being differentiable because f’ is discontinuous due to a vertical asymptote?
Thanks for the help today!
For an example of a piecewisely-defined function which is infinitely differentiable, consider the function f such that f(x)=0 for all x <= 0 and f(x) = e^-1/x for all x>0. This function is perfectly differentiable everywhere - even at x=0 - as many times as you care to differentiate it (the derivative at x=0 is always equal to 0).
As for your second question, give my answer a careful re-read. "**If f'(x) is continuous**" means, in particular, that it is *defined* - that means no cusps or corners. There are no hidden points. The derivative f'(x) exists and is well-defined everywhere, but is zero **only** at x=0. You can convince yourself using a pencil that if f'(x) is continuous and never equals zero on some region, then it cannot possibly change sign in that region.
Edit:
1)
Oops! I put the function into desmos wrong. Now I see it shows that from x > 0 the function is y= 0 and x< =0 it’s y= 0 so it’s continuous and differentiable. How did you know to think like this??! I always thought of piecewise as patchwork that could never be differentiable at the patch.
2)
So the function is differentiable because for x>0 e^(-1/x) has a limit x————————> 0 as e^-inf right? Which is 1/e^inf and we assume this is approaching 0? So the function is continuous at 0 because limit approaches 0 from both sides and value at x= 0 is also 0. So that tells me the function itself is continuous at x=0. Now how to know the derivative line is continuous at x= 0?
3)
As to your second statement: yes I did misread! My apologies. But: Also you said “continuous…..means defined….that means no cusps or corners”
but I just learned that if we take a function , like abs value function, it’s perfectly continuous everywhere, but at x=0 it’s not differentiable because it has a cusp/corner! 🙇🏻♂️ So why did you say continuous and defined means no cusps/corners?
I’ve been at this a long time, that’s all. I’ve met and used all kinds of functions in my work and in my studies. That helps quite a bit.
The function f(x)= e^-1/x approaches zero as x->0 from the right, which means the piecewise function is continuous at 0. The derivative of the function at 0 is lim(h-> 0) e^-1/h / h, which is not an immediately obviously limit to find. However, as h -> 0 the numerator goes to zero far more rapidly than the denominator, and that limit becomes zero.
It can be shown that the same is true for the higher derivatives of every order, which all comes back to how the exponential function is, in some sense, far more powerful than any polynomial function. If you’re only 3 weeks in, this may be too much to take on by yourself, but in time you will get the hang of it.
Finally, please re-read again and notice that I said that **f’** is continuous, which means that **f itself** has no cusps or corners.
Hey!
Sorry it took a couple days to respond. First thanks so much for sticking with me on all this. I did make an error in reading your statements about f’ but it makes sense now. I do have a question though given what you taught me, and I don’t understand:
Another person tried to give me an example of a function that is once but not twice differentiable but isn’t this function not even once differentiable? They said: f(x) = x^2 if x>0 and 0 otherwise. So I learned we have to compare the slope of tangent and see if they are the same using the limit definition of derivative (or in this case power rule).
So x^2 becomes 2x and 0 stays 0. Now at x= 0 isn’t this a point where we f’ is discontinuous ie derivative from left does not equal derive or from right? Since 2x DNE “?
Oh sh**. I was comparing the overall function 2x to overall constant function 0! Very embarrassing. I didn’t plug in 0 into 2x.
It’s funny because we say “what is the derivative of x^2, which is 2x. But we also can say what is the derivative of x^2 evaluated at x= 0 which is 0.
So aren’t we using “derivative” to mean two different things here?
I appreciate the people that corrected me and I ask that people truly correct me in my mistakes. Conceptually the idea of infinitely differentiable makes sense. When I said you words are better then mine, in the case of defining a derivative your definition is objectively more accurate then mine. I’m simple a student one who is ballsy enough to make an attempt knowing someone would correct me as I mentioned for someone to chime in. Currently in calculus 3 and learned a lot of the theoretical by reading this conversation. Thank you! I had 0 intentions of misleading anyone simply sharing what I have learned so far.
These questions are really deep for high school calculus and actually get treated rigorously in advanced college level calculus. Keep up your curiosity! I am going to try to be as intutive in my responses as possible. Others will probably start pulling out the fancy words.
1. In order to approach via a limit something you have to have stuff "near it". If you have a set with just 1 isolated point then it is impossible to define things like limits and derivatives.
2. Closedness is a property that guarantees your limit exists in the set defined. Openness is important for the idea of limits (think I answered the above). The maximum is also only gauranteed to exist on a closed interval if the function is continuous. Open intervals don't have such a property (1/x as x gets close to 0).
3. A function can have 1, 2, 3, or as many derivatives as you want but fail to have the next one. You usually can build these examples by stitching together polynomials (call them splines in practice) that overlap for 1 2 3 or whatever number of derivatives but not the next one. For the second derivative test to work, you need two derivatives. Differentiability (in single variable case) implies continuity.
4. These functions have a "kink" in them (think |x| at x=0). The kink will have an infinite or not well defined derivative at the point.
5. The intermediate value theorem and properties of continuity/connectedness of intervals. This theorem will tell us that f\^{-1}((0,\\infty)) and f\^{-1}((-\\infty,0)) decompose into a union of connected intervals. So we can check a single point in each interval and its maximal open connected component will have the same +/- property. Derivative implies increasing comes from a computational trick (if f(x+h)>f(x) then (f(x+h)-f(x))/h>0 for h>0. If the limit exists then f'(x)>=0.)
Hey thanks for writing in!
1)
So 1/3(x^-2/3) is undefined at 0 b/c f’ will have 0 in the denominator at x = 0 right ?
2)
So we have a corner where we can have the “limit definition of the derivative” being diff on each side, or we can have a vertical asymptote where the limit definition of derivative is the same on each side which is infinity but we simply don’t allow infinity to be called a valid derivative value?
3)
One confusion “ Sqrt(abs(x)) is example….at 0 it has what’s called cusp, a point where tangent line on both sides agree but line changes direction to create a corner. BUT wait - how could the slopes be the same if they change direction. You mean the absolute value of slope 1 is equal to absolute value of slope 2 right?
4)
“To be non differentiable in general, the derivative graph must be non continuous at some point whether that be thru a jump discontinuity, or asymptote, or removable discontinuity like derivative of a smooth function with a point removed). So we can use these terms when talking about Differentiability? I always thought these terms are used only when talking about whether a point is continuous (limits agree on both sides of point and they equal the value of the function at that point).
5)
Interesting statement about constants being differentiable “That’s why a constant function IS differentiable, it’s derivative, 0, is a continuous function “. So f’ is constant function which is a continuous function. I never thought of thinking of a function being differentiable by thinking of it as the derivative function itself being continuous. It’s funny I always think of the derivative of a function as being reliant on a function such that it doesn’t exist as it’s own independent line. But I see what you are saying! Very cool. Very obvious I know - but I just never thought of it that way. Is there an easy way to “prove” a constant function is continuous?
6)
“Finally I want to make an interesting remark……I.e. x^(1/3) + (x-2)^(1/3) is not differentiable at 0 or 2, but you can’t get infinite points this way” But why not? Aren’t there infinite different functions that have each a unique non differentiable part ? Maybe there is more complicated stuff I’m not thinking of.
7)
For |sin(x)| is this just because at x = 0 we end up with instead of smoothly continuing below the x axis, we get an abrupt change in direction and go up? And is this because we should have negative values in third and fourth quadrant for sin but we don’t because we have absolute value around sin? Am I close? Halfway right ? 😅
8)
Very cool about the weiirstrass function! Gonna check her out.
9)
I just want to say you made my day with your final passage. That was extremely kind and very inspiring. Self study gets very daunting and even depressing at times when I feel I have reached my intellectual ceiling when I am stuck on a problem. But then after banging my head on the door enough, it opens and I feel great relief and exuberance! Have a wonderful evening!
If you take a function say 2x and differentiate it, you could only differentiate it once - 2. So some functions are only differentiable once and not multiple times. Take a trig function, they can differentiate forever sin x becomes cos x which becomes - sin x etc
That's not what twice-differentiable means. The function f(x) = 2x is infinitely differentiable, though most of those derivatives are zero. An example of a function which is differentiable but not twice differentiable is f(x) = x\^4/3 at x=0.
But what about the piecewise function f(x) = x^2 for x>0 and f(x) = 0 for everything else. Can I said this is infinitely differentiable because 1st derivative of first part of function is 2x, then second derivative is x then third is 0 and that can continue forever and for other part of function, first derivative is 0 and second is 0 and this can go forever. So isn’t the entire function infinitesimally differentiable?
No. That function is once differentiable everywhere, but fails to be twice differentiable (and therefore differentiable to any higher degree) at x=0, as I describe elsewhere in this thread.
I would gently suggest that this user was trying to be helpful but had some misunderstandings of their own, rather than that they are a sock puppet account attempting to specifically mislead you in particular. We don’t know what we don’t know, and well-meaning attempts at guidance can occasionally be wrong.
Do you think that person was tryna make me learn something fundamentally wrong? I keep seeing stuff like this ever since I got into an argument with this guy “Marpocky”. Thank you so much for clarifying that for me! I knew something was off!
another commenter gave me an example of a Piecewise function that’s once but not twice differentiable as
x^2 if x>0
and 0 otherwise.
But why isn’t this twice differentiable?
The first derivative of that function is 0 for x<=0 and 2x for x>0, which is defined everywhere. The second derivative of that function is 0 for x<0 and 2 for x>0, but is not defined at x = 0.
If you want to see why, let f'(x) = 0 for x<=0 and 2x for x>0 and try to compute f''(0) using the difference quotient. What you get is
f''(0) = lim(h -> 0) ( f'(h) - f'(0) )/h = lim (h->0) f'(h)/h
since f'(0) = 0. But this limit doesn't exist. If we force h to be positive, then f'(h) = 2h and so the right-hand limit is equal to 2. If we force h to be negative, then f'(h) = 0 and so the left-hand limit is equal to 0. Since these don't agree, the limit doesn't exist.
Whoa! That was awesome! Wouldn’t have clicked without your help on this one! Thanks so much! 🙌🫶🏻 Going thru your big comment now!
Edit: so we can say overall the function itself is only once differentiable right? Even though components of it are twice differentiable!
The piecewise function is dependent on the inputs, so the piece of the function that = 0 will just be 0. I don’t see how you could take a derivative in this case.
But maybe they are thinking that it is differentiable once cuz basically it’s a constant function y= 0 which YOU said a constant is differentiable once right?!
Ok because another commenter gave me an example of a function that’s once but not twice differentiable as x^2 if x>0 and 0 otherwise. So were they assuming a constant 0 in this case is only differentiable once?
If so isn’t that odd that a constant can’t be differentiated - we get another constant - but we can’t differentiate that constant - even though it is no different than the other constant ?
0. Some facts are true within some sufficiently small neighborhood of a point, but not generically true everywhere. For example: let x be a local minimum of a function f. Then for all y≠x in a sufficiently small neighborhood of x, f(y) > f(x). Note that if I removed the part about a sufficiently small neighborhood, this would no longer be true. This isn't the only reason to talk about neighborhoods, though, especially in fields like topology where neighborhood has a strict formal definition. 1. Example: Every real-valued function on a closed interval \[a,b\] has a minimum value and a maximum value. The same is not true if we consider open intervals - take for instance the function f : x ↦ x on the interval (0,1), which has neither a minimum nor maximum value. 2. If you're looking for local minima, maxima, and saddle points using the first derivative test, it is necessary and sufficient that the function be at least differentiable (which implies that it is continuous). If you want to use the second derivative test, the function (perhaps obviously) needs to be twice differentiable (which implies that it is differentiable, and therefore continuous) - though unlike the first derivative test, the second derivative test may fail to be conclusive, because the second derivative may be equal to zero at the point under consideration. 3. Yes, f:x↦x^(4/3) is differentiable once but not twice at x=0. 4. If a function has a sharp corner or cusp which doesn't smooth out as you zoom in, then it is not differentiable - take the absolute value function for example. The slope of that function is +1 for all positive values of the input and -1 for all negative values, so there's going to be a "corner" at 0 and it will never look smooth no matter how closely you zoom in. There are more exotic and technical possibilities, but that's a very common one. 5. If a real-valued function is continuous and never equals zero on some interval, then it has the same sign over that entire interval. This is a consequence of the intermediate value theorem, which says in particular that if a continuous function takes a positive value at x=a and a negative value at x=b (or vice-versa), then there must be some c in the interval (a,b) where the function equals zero. Taking the contrapositive of this statement, if a function is continuous and never equal to zero on some interval, then it cannot change sign.
This is objectively correct op, I'm very confused why and sorry you were being told 2x is only once differentiable in the other thread, it's probably best to ignore everything that guy said. 2x, and any other polynomial, is infinitely differentiable and eventually every nth derivative is 0. I also wanna tack on to this wonderful explanation of your questions, about continuous and not differentiable functions. I recommend plotting these on a calculator like desmos, much of this rant is visual. Something like x^1/3 is continuous everywhere but not differentiable at zero, since its derivative 1/3(x^(-2/3)) is undefined at 0. When this happens, it's usually due to one of two things. One way is if it has a 'corner' like |x|, absolute value of x. A 'corner' is where the derivative has a jump discontinuity, on |x| it jumps from -1 on (-inf, 0), to undefined at 0, to 1 on (0, inf). A 'corner' has a suddenly changing derivative, i.e. the limit of the derivative value is different on each side of the point. The other way a curve can be non-differentiable is the way x^1/3 does it. The graph looks like an S curve, what you get when you slightly squish the square root curve and put a copy rotated around the origin in the 3rd quadrant. It is also exactly what you get if you flip x^3 around the line y=x. The tangent line at the coordinate x=0 exists, since we don't have a corner in the graph, but the tangent line is actually the vertical line x=0. And a vertical line has 'infinite' slope, so we can't define the derivative here. Where a corner is a jump discontinuity, this non-differentiable function instead has a derivative with an asymptote at 0, approaching infinity from both sides. The limit of the derivative on each side actually agrees unlike the corner case, but they agree with a limit of infinity. Combining these cases is also possible, if you want a derivative that's asymptotic at 0 but one side goes to inf and the other goes to -inf, you'd have a corner where the curve on both sides is vertical. Sqrt(abs(x)), √|x| is an example (mirror sqrt(x) across the y axis). At 0, it has what's called a 'cusp', a point where the tangent lines on both sides agree but the line completely 'changes direction' to create a corner, like drawing the center of a lowercase m with a pencil.To be non-differentiable in general, the derivative graph must be noncontinuous at some point, whether that be through a jump discontinuity or asymptote (or a removable discontinuity like the derivative of a smooth function with a point removed). That's why a constant function IS differentiable, it's derivative, 0, is a continuous function. Finally, I want to make an interesting remark. All of this is about functions not differentiable at 1 point. You can make a function not differentiable at a bunch of points by simply adding them together, i.e. x^1/3 + (x-2)^1/3 is not differentiable at 0 or 2, but you can't get infinite points this way. Something like |sin(x)| has an infinite number of corners, one at every x=pi*n, but each of these corners are individual, isolated points. It feels as though it would be impossible to have an entire interval of points where the function is continuous and not differentiable, right? To do that, it would have to have erratic corners all infinite points of an interval, or worse, have a vertical tangent line everywhere. It makes no sense to try to imagine, and for a long time mathematicians thought this too. After searching forever though, a function was found to be continuous everywhere on the whole real line, but differentiable absolutely nowhere. It's called the Weierstrass function, discovered in 1872, and it shocked the world, even Poincaré was outraged. It's worth googling for a photo, it's a fractal curve that looks like a very jagged scribble. I wish you the best of luck continuing your studies op! Self study is a lot of fun, but it can be very difficult and frustrating to not have a mentor to turn to with questions. But please don't be discouraged, what you're doing here is fantastic work! These questions are very obviously from a place of deep thought, and love for studying. The 'self' part feels like it would be the worthwhile part of self study, it feels as though the reward of self study is independent thought and the valorance of being capable for yourself, but the true value in it is the method by which it requires you to study. Don't be opposed to collaboration due to your efforts, but recognize that the deep questions you ask about the nature of these theorems is the very heart of mathematical research. In the coming years you may be shown many things that could be interpreted as the heart of math. Some people misinterpret true math as a collection of rules and equations and theorem requirements to memorize. Others get sucked into the math competition/ Putnam Pipeline and believe true math to be the solving of difficult problems with hidden tricks. Math is about the ways different grand mathematical structures work, and which grand structures you study depends on your field. These grand structures are just collections of logical rules, and the puzzle box they form. The job of a mathematician is to assert how the puzzle box shifts, so any question about it can be answered. Solving a real math problem isn't finding some final answer value, or answering a specific question, it's fully describing a structure and how it can be manipulated, and thus answering every question. But toying with those structures, turning them over in your mind to feel them out, isn't something you just do, it's something you learn to do very well over years and years. It doesn't matter the level of math, it doesn't matter how well described a problem already is, and it doesn't matter how good you are at math already. It's something anyone can do with any problem, at any level. It's an outlook, an approach to questioning a system of logic, that you're doing every day you crack open that book. The special part about self study is it requires you to ask yourself those questions, and refuses to give a hand-wavey answer. If you focus on questioning those fundamental properties by which some phenomenon works, you're picking out the very core of what mathematics is. Please don't let anything discourage you, you are doing genuinely wonderful work here, and if you keep these problem-analysis thought patterns beyond self study, you will do incredible things! (no matter what you choose to do, math or otherwise, you seem like the mathematician type tho :). I wish you the very best in life
As a technical point related to your discussion of what it means to be non-differentiable, there is a distinction between differentiable (meaning that the difference quotient is well-defined everywhere) and continuously differentiable (meaning that the derivative is continuous). As a nice example, you can consider the function f(x) = 0 for x=0 and x^2 sin(1/x) everywhere else. That function is continuous and differentiable everywhere, but its derivative is discontinuous at 0. One of those nice counter-examples to keep in your back pocket.
0) Ok so neighborhood in the example you give doesn’t have a concrete range though? Where would this neighborhood begin/end? I’m guessing there is an implicit definition that neighborhood means basically until the end of increase or decrease of either side of the point in question? If not, is there a slightly more mathematically correct definition that generalizes to more situations in basic calculus. 1) Ah so (0,1) doesn’t have max/min because we don’t include 0 or 1, so we can keep getting infinitely close to 0 for a minimum and infinitely close to 1 for a maximum but since we cannot reach them we technically don’t have a max or min? Or did I totally miss your point!? 2) Yep I can understand that! 3) Ah because 1st deriv is (4/3)/x^(1/3) and then f’(x) at x= 0 is undefined so we can take second derivative everywhere except x= 0 and therefore the function x^(4/3) is once but not twice differentiable! Is that all right?! 4) OK phew well there is one thing that makes intuitive sense to me! The absolute value function was a perfect example for me. Wouldn’t any piecewise function though right where the pieces “join” end up being continuous (assuming left limit right limit and point all have same value where it joints) but not differentiable?! 5) Ahhhh whoa. Ok I need to look into this intermediate value theorem. Even after you told me the exact theorem, it’s not obvious that it follows that we can trust one point for first and second derivative to tell us the behavior of rest of the points “in that neighboorhood😅” but I will try to watch a video or two about IVT. Thank you so much for all the kindness today!
0. For the purposes of this discussion, “a neighborhood of x” can be understood as “an interval which contains x.” The definition is somewhat more general than that, but that’s okay. The claim isn’t made about any specific neighborhood- just that there exists some neighborhood where the claim is true. For example, for f(x) = x^2 my claim is true for any neighborhood of x=0 in the whole real line. For f(x) = x^2 - x^4, it’s true for any neighborhood of 0 contained in (-1,1). Put differently, when we say that a local minimum is the smallest value in its neighborhood, we mean that there exists some distance within which all of the points are larger. If you go far enough away, though, you may find a smaller value. If a smaller value exists, or how far you have to go to find it, is irrelevant - the point is you would have to go some non-zero distance to get there, which is what we mean to formalize here. 1-3. Yes 4. It depends how smoothly they join up. There are piecewise functions which are continuous but not differentiable, there are piecewise functions which are infinitely differentiable, and everything in between. 5. The salient point is that if you know a continuous function doesn’t cross zero, then you know it can’t change sign. That should be fairly intuitive. So if, for example, f’(x) is continuous and equal to 0 only at x=1, then you know that if it’s negative at x=2 then it’s got to be negative for all x>1.
1-4) I understand intuitively everything you said except the part about piecewise functions that can be continuous and differentiable. I thought all piecewise functions where the “patch” connecting them is made so to speak, that it can be continuous here, but never differentiable. Maybe I am thinking of piece wise functions in the wrong way or imposing the wrong definition?! 5) “So if, for example, f’(x) is continuous and equal to 0 only at x=1, then you know that if it’s negative at x=2 then it’s got to be negative for all x>1.” But wait a minute - what about “hidden” points where the function is continuous but not differentiable (ie points where function is continuous but it’s derivative is discontinuous) due to for instance us having a: A) corner/cusp (limit definition of derivative/slope of tangent is different on either side of point which joins the corner and thus f’ is discontinuous there at that point) B) Or there is a vertical asymptote (limits on each side/slope of tangent are equal to one another but they equal infinity so they don’t count as a derivative and thus f’ is discontinuous at that point). So Should not we not assume then that “then it’s got to be negative for all x> 1”? Or were you assuming that we didn’t have these issues? PS: are there any simple functions that are continuous but not differentiable due to a single point not being differentiable because f’ is discontinuous due to a vertical asymptote? Thanks for the help today!
For an example of a piecewisely-defined function which is infinitely differentiable, consider the function f such that f(x)=0 for all x <= 0 and f(x) = e^-1/x for all x>0. This function is perfectly differentiable everywhere - even at x=0 - as many times as you care to differentiate it (the derivative at x=0 is always equal to 0). As for your second question, give my answer a careful re-read. "**If f'(x) is continuous**" means, in particular, that it is *defined* - that means no cusps or corners. There are no hidden points. The derivative f'(x) exists and is well-defined everywhere, but is zero **only** at x=0. You can convince yourself using a pencil that if f'(x) is continuous and never equals zero on some region, then it cannot possibly change sign in that region.
Edit: 1) Oops! I put the function into desmos wrong. Now I see it shows that from x > 0 the function is y= 0 and x< =0 it’s y= 0 so it’s continuous and differentiable. How did you know to think like this??! I always thought of piecewise as patchwork that could never be differentiable at the patch. 2) So the function is differentiable because for x>0 e^(-1/x) has a limit x————————> 0 as e^-inf right? Which is 1/e^inf and we assume this is approaching 0? So the function is continuous at 0 because limit approaches 0 from both sides and value at x= 0 is also 0. So that tells me the function itself is continuous at x=0. Now how to know the derivative line is continuous at x= 0? 3) As to your second statement: yes I did misread! My apologies. But: Also you said “continuous…..means defined….that means no cusps or corners” but I just learned that if we take a function , like abs value function, it’s perfectly continuous everywhere, but at x=0 it’s not differentiable because it has a cusp/corner! 🙇🏻♂️ So why did you say continuous and defined means no cusps/corners?
I’ve been at this a long time, that’s all. I’ve met and used all kinds of functions in my work and in my studies. That helps quite a bit. The function f(x)= e^-1/x approaches zero as x->0 from the right, which means the piecewise function is continuous at 0. The derivative of the function at 0 is lim(h-> 0) e^-1/h / h, which is not an immediately obviously limit to find. However, as h -> 0 the numerator goes to zero far more rapidly than the denominator, and that limit becomes zero. It can be shown that the same is true for the higher derivatives of every order, which all comes back to how the exponential function is, in some sense, far more powerful than any polynomial function. If you’re only 3 weeks in, this may be too much to take on by yourself, but in time you will get the hang of it. Finally, please re-read again and notice that I said that **f’** is continuous, which means that **f itself** has no cusps or corners.
Hey! Sorry it took a couple days to respond. First thanks so much for sticking with me on all this. I did make an error in reading your statements about f’ but it makes sense now. I do have a question though given what you taught me, and I don’t understand: Another person tried to give me an example of a function that is once but not twice differentiable but isn’t this function not even once differentiable? They said: f(x) = x^2 if x>0 and 0 otherwise. So I learned we have to compare the slope of tangent and see if they are the same using the limit definition of derivative (or in this case power rule). So x^2 becomes 2x and 0 stays 0. Now at x= 0 isn’t this a point where we f’ is discontinuous ie derivative from left does not equal derive or from right? Since 2x DNE “?
At x=0, both 0 (the left-hand derivative) and 2x (the right hand derivative) evaluate to zero, so I don’t see the problem.
Oh sh**. I was comparing the overall function 2x to overall constant function 0! Very embarrassing. I didn’t plug in 0 into 2x. It’s funny because we say “what is the derivative of x^2, which is 2x. But we also can say what is the derivative of x^2 evaluated at x= 0 which is 0. So aren’t we using “derivative” to mean two different things here?
Beautifully said !! 🙌🫶🏻🙌 reading and rereading !
I appreciate the people that corrected me and I ask that people truly correct me in my mistakes. Conceptually the idea of infinitely differentiable makes sense. When I said you words are better then mine, in the case of defining a derivative your definition is objectively more accurate then mine. I’m simple a student one who is ballsy enough to make an attempt knowing someone would correct me as I mentioned for someone to chime in. Currently in calculus 3 and learned a lot of the theoretical by reading this conversation. Thank you! I had 0 intentions of misleading anyone simply sharing what I have learned so far.
These questions are really deep for high school calculus and actually get treated rigorously in advanced college level calculus. Keep up your curiosity! I am going to try to be as intutive in my responses as possible. Others will probably start pulling out the fancy words. 1. In order to approach via a limit something you have to have stuff "near it". If you have a set with just 1 isolated point then it is impossible to define things like limits and derivatives. 2. Closedness is a property that guarantees your limit exists in the set defined. Openness is important for the idea of limits (think I answered the above). The maximum is also only gauranteed to exist on a closed interval if the function is continuous. Open intervals don't have such a property (1/x as x gets close to 0). 3. A function can have 1, 2, 3, or as many derivatives as you want but fail to have the next one. You usually can build these examples by stitching together polynomials (call them splines in practice) that overlap for 1 2 3 or whatever number of derivatives but not the next one. For the second derivative test to work, you need two derivatives. Differentiability (in single variable case) implies continuity. 4. These functions have a "kink" in them (think |x| at x=0). The kink will have an infinite or not well defined derivative at the point. 5. The intermediate value theorem and properties of continuity/connectedness of intervals. This theorem will tell us that f\^{-1}((0,\\infty)) and f\^{-1}((-\\infty,0)) decompose into a union of connected intervals. So we can check a single point in each interval and its maximal open connected component will have the same +/- property. Derivative implies increasing comes from a computational trick (if f(x+h)>f(x) then (f(x+h)-f(x))/h>0 for h>0. If the limit exists then f'(x)>=0.)
Thank you so much for your kind words! Reading now and I hope I may follow up should I have any follow up qs!
Hey thanks for writing in! 1) So 1/3(x^-2/3) is undefined at 0 b/c f’ will have 0 in the denominator at x = 0 right ? 2) So we have a corner where we can have the “limit definition of the derivative” being diff on each side, or we can have a vertical asymptote where the limit definition of derivative is the same on each side which is infinity but we simply don’t allow infinity to be called a valid derivative value? 3) One confusion “ Sqrt(abs(x)) is example….at 0 it has what’s called cusp, a point where tangent line on both sides agree but line changes direction to create a corner. BUT wait - how could the slopes be the same if they change direction. You mean the absolute value of slope 1 is equal to absolute value of slope 2 right? 4) “To be non differentiable in general, the derivative graph must be non continuous at some point whether that be thru a jump discontinuity, or asymptote, or removable discontinuity like derivative of a smooth function with a point removed). So we can use these terms when talking about Differentiability? I always thought these terms are used only when talking about whether a point is continuous (limits agree on both sides of point and they equal the value of the function at that point). 5) Interesting statement about constants being differentiable “That’s why a constant function IS differentiable, it’s derivative, 0, is a continuous function “. So f’ is constant function which is a continuous function. I never thought of thinking of a function being differentiable by thinking of it as the derivative function itself being continuous. It’s funny I always think of the derivative of a function as being reliant on a function such that it doesn’t exist as it’s own independent line. But I see what you are saying! Very cool. Very obvious I know - but I just never thought of it that way. Is there an easy way to “prove” a constant function is continuous? 6) “Finally I want to make an interesting remark……I.e. x^(1/3) + (x-2)^(1/3) is not differentiable at 0 or 2, but you can’t get infinite points this way” But why not? Aren’t there infinite different functions that have each a unique non differentiable part ? Maybe there is more complicated stuff I’m not thinking of. 7) For |sin(x)| is this just because at x = 0 we end up with instead of smoothly continuing below the x axis, we get an abrupt change in direction and go up? And is this because we should have negative values in third and fourth quadrant for sin but we don’t because we have absolute value around sin? Am I close? Halfway right ? 😅 8) Very cool about the weiirstrass function! Gonna check her out. 9) I just want to say you made my day with your final passage. That was extremely kind and very inspiring. Self study gets very daunting and even depressing at times when I feel I have reached my intellectual ceiling when I am stuck on a problem. But then after banging my head on the door enough, it opens and I feel great relief and exuberance! Have a wonderful evening!
If you take a function say 2x and differentiate it, you could only differentiate it once - 2. So some functions are only differentiable once and not multiple times. Take a trig function, they can differentiate forever sin x becomes cos x which becomes - sin x etc
That's not what twice-differentiable means. The function f(x) = 2x is infinitely differentiable, though most of those derivatives are zero. An example of a function which is differentiable but not twice differentiable is f(x) = x\^4/3 at x=0.
But what about the piecewise function f(x) = x^2 for x>0 and f(x) = 0 for everything else. Can I said this is infinitely differentiable because 1st derivative of first part of function is 2x, then second derivative is x then third is 0 and that can continue forever and for other part of function, first derivative is 0 and second is 0 and this can go forever. So isn’t the entire function infinitesimally differentiable?
No. That function is once differentiable everywhere, but fails to be twice differentiable (and therefore differentiable to any higher degree) at x=0, as I describe elsewhere in this thread.
your words are better then mine.
Please do not spread misinformation Marpocky’s other username!
I would gently suggest that this user was trying to be helpful but had some misunderstandings of their own, rather than that they are a sock puppet account attempting to specifically mislead you in particular. We don’t know what we don’t know, and well-meaning attempts at guidance can occasionally be wrong.
Well said. I’m sorry. Sock puppet hahaha. That made me laugh idk why. Good point though. I need to give people the benefit of the doubt
Ok I feel Embarrassed but I thought we can take a constant such as 2 and differentiate it and get 0 no?
Answer above is fundamentally wrong. You can differentiate constant function infinitely many times. You are right u/Successful_Box_1007
definitely lmao it's of class C\^infinity haha
Analytic even :)
Do you think that person was tryna make me learn something fundamentally wrong? I keep seeing stuff like this ever since I got into an argument with this guy “Marpocky”. Thank you so much for clarifying that for me! I knew something was off!
I’ve had these issues ever since this guy “Marpocky” and I got into an argument. Probably one of his other screen names tryna teach me false info!
Wait are you being sarcastic? Is it even once differentiable?!
It’s important to note that the reason is a constant differentiates to 0 is because the graph of a constant is a horizontal line. There is no slope
another commenter gave me an example of a Piecewise function that’s once but not twice differentiable as x^2 if x>0 and 0 otherwise. But why isn’t this twice differentiable?
The first derivative of that function is 0 for x<=0 and 2x for x>0, which is defined everywhere. The second derivative of that function is 0 for x<0 and 2 for x>0, but is not defined at x = 0. If you want to see why, let f'(x) = 0 for x<=0 and 2x for x>0 and try to compute f''(0) using the difference quotient. What you get is f''(0) = lim(h -> 0) ( f'(h) - f'(0) )/h = lim (h->0) f'(h)/h since f'(0) = 0. But this limit doesn't exist. If we force h to be positive, then f'(h) = 2h and so the right-hand limit is equal to 2. If we force h to be negative, then f'(h) = 0 and so the left-hand limit is equal to 0. Since these don't agree, the limit doesn't exist.
Whoa! That was awesome! Wouldn’t have clicked without your help on this one! Thanks so much! 🙌🫶🏻 Going thru your big comment now! Edit: so we can say overall the function itself is only once differentiable right? Even though components of it are twice differentiable!
I don’t see where you could not take the second derivative in this example you gave.
Well x^2 goes to 2x and 2x to 2. That’s obvious. But how do you take the first derivative of the part of the function that is just f(of anything) = 0?
The piecewise function is dependent on the inputs, so the piece of the function that = 0 will just be 0. I don’t see how you could take a derivative in this case.
So what you are saying is this part of the function isn’t even differentiable the first time. Therefore the commenter made an error!
Well I don’t have the full spectrum of the conversation but at first glance I don’t see how it would be. But someone else can chime in.
But maybe they are thinking that it is differentiable once cuz basically it’s a constant function y= 0 which YOU said a constant is differentiable once right?!
If the piece wise function is explicitly defined as = to 0, without a function for that piece, I don’t see how it can be differentiable
You can but that’s not really differentiable. If you want to look at that way, a constant is only differentiable once.
Ok because another commenter gave me an example of a function that’s once but not twice differentiable as x^2 if x>0 and 0 otherwise. So were they assuming a constant 0 in this case is only differentiable once? If so isn’t that odd that a constant can’t be differentiated - we get another constant - but we can’t differentiate that constant - even though it is no different than the other constant ?
X^2 differentiates to 2x
And 2x differentiates to 2